'''' ' ɇ 'ՏçՏ ɇ £ ":e4> ΛՏՏ ƃɇ ɇ £ gńĭ ń&ń'ɇ ťłĉɮ '\ń ŦƋŃĊɮ [ɇ ɇ £ 'Eń£ ńƚɇ ɇ ɗǔՏ
bûÛń8_ń2 ńNPń*ńơɮ ń,8ńQ}p.x.ZX;X *ń7#ń!ńi+ńĨń yO.aFX<~&! /,^ń ńµ7¯ńń / %!$.4 ¬N ń*ń*ÎoĆ ń#8Fń . ń Ƥɮ ńP^S[.v ·¶,ń #®7ńń 22ń ¼*ńH! ńH. #nɮ
Eń
nɮ ! ,Fń /_ń ,7\ń 8] F bÂ
ńJ. Êȷ вӧՏ n.v.Y\:Y j Ĥɇ
Ƅɇ
ħɇ
.ĩń@ń1: ĶÜń . Stń 5ùń@ń°Տ TՏ /6 ń I=v=Ekx: k88 dK ń ĕՏ ɇ ĖՏ ɖՏ Eń xՏ +Տ G>x> ҸžՏr Ť: }ĚrQ ń `HńӉՏ \Տ Տ ńâćɮ ĪzńĕՏ \Տ ȷ _l6 .e8 īě:ĜńҮՏ B ZǦɐ xՏ Տ ɮ ĻÝńгӨՏ s
tÕOń `ïńӊՏ \Տ TՏ ē ń GĆjȷ Ĭçń wՏ Ėɇ (~ `m6 ƀá { ń
ńwՏ ń Z8
@ªń ſՏ (2ńýQ}
þÖOèń d
ńsɇ ėɇ ȷ ² Ē ńãĈɮ Uu ńj ńǶɇ C ič ńKÞńҭՏ ®Տ ʊ ńwՏ ń©ǕՏ e6 ń@ńՏ ʉՏ R~ 5KńđPDńĤ6ßń~üQń
ʋՏ
£Ǚ
Êń<jĹńÒ2Տ 8ąՏՏ rɇ P-'B'S ƶɇ
B
ǞɮŘՏ ǥ Տ
Å0 0Iń
Ƽ&ŖǤưɮ 'ńȷ ȥɐ xՏ ' %" ɕͼՏ A
ű?ɮZ
ŰȬɮ
ȭɮՏ
/Տ Տ c?
Տ ǑՏ
ńń;0
¿× +ńLÏń"ñń' ˚ cöɮ ń sĢŢ®Z aÚɮ ˙Տ ÷ɮ [ X ȮɮՏ Տ ljɐ sģȑɮ ŚՏ /Տ 1Տ ?
f:Ań
ȸǐՏ
ńńń>;BZ
=%ńńń=m; Zń c<ń
Ǜ
.ńń =ń Vȷ #Y XɮŘÿɮ9Āɮ ! Ëń÷ĝń$!ńÐń1 ń 0 į%ńhȷELM 0'9GS fLńIġń B%ń!ÿńúńSğI;ńĉ ©ń ɍ Wɮďɮř.ɮ ƞāɮ̅ŻՏ ƨ.ɮ !.ɮ9ĂɮƪɎɮ9% ɌɮȜżՏ ! "JVńńęķĀ 5ń 'Sv?ń$ĽńėĞ+ńńɮ $ ńC'S ńõáqń K??<E'S é ńń$5Vń< ń $ Ań À" ń+ ľńxń :ń5ń5'EGS <9 'S G.'@'S 2ES S GĠńļń!ń1ńĮńĊ =ńń čɮĎɮ v ń °ńVń3ń?ĥń5 Ŀń:A 0 ń$A[ń ÆńP'S /O'S ˤՏ ?5R'#S ńɮ%'ń"ńń 'ń Bń9#S +> ń+ń Ďńń 'ń ¾Uń ¦Wń´ńŇjȷ)ȷ ĦńÑO?ńSðyń ń5<ń 'ń ń > ń ńā ń LêńBńń<:5QS "3%ńUń^Տ Ô ńɮ C'S ńń'Ăàń ½ń>ń$ńĕ%"Øń1ńG.'8S ĔńĖ 1 3ń1 ńmGń ńń'S (DEIS !#' $"' ńx"yń%ń'
'
ljȂȝǹǠȪȢȯwɮ ǂ>ȶǽǯȗ>ȃǩ>Ȕɮ NJ Ҥ ǪȄwɇ ±
1Q\D7f_h@ (7sF@[7xR;7S *S~ZfQ7? 1(
Þɮ
1B\Q_j 1 C9rQ_\
ƑƓ5Į ŧŷƔ
Ǎǫe Vɀøɐ ɐƸȮȓ[ɐ Ŷ14
$Zf_hr7\r
t¡½KÆ ++Æ Æ ³K¢bzt¢Æ %tKÆÂz³Æ 1t¢¾K¢Æ ztÆVKÆ 1t¢¾KÆ £VKKƹ_JKJÆ 'zƬVKÆo³j_jOÆ @Zz_>KÆ ³O_ztÆ KvKÆÃz³Æ 3u¢¾LÆ ztÆZKÆ 1u¢¾MÆ¢VKKÆ<ÃÆ¢W4JcuRÆ ¬[NÆ=´99jKÆ Gzr²6_yfrSÆ XOÆ kK±KÆ Æ Æ *Æ ½Տ zÆ ƲÑ >}zuIdtQÆ zÆ ®WKÆ FzDÆ 1u¢¿K )zƯVKÆ z°\NÆ¡Vz¯Æ ·K¤bzt¢ Æ ¿KÆÂz´Æ1u¢¾NÆ zuÆ]KÆ 1u¢¾KÆ¢WKK§Æ 1uIÆ¢Z1IKƯ]KÆ 1e1KÆ 9³9=nK¢Æ 9Kjz¾ÆÄz¶Æ 1u¢»KÆ .zÆ ¢K
¢Æ ţǹɇ wKKIKIÆzÆh³¢_Ǽɐ Äz·Æ 5x¡¾K¥ &1E^Æ µK¢eztÆ F1dK¢Æ ɁՏ ѣӕхȄՏ .zÆ H3j?¸j7~Æ 8Æ 8jjzºOIÆ
+'!1" ̺̄Տ ( ( ( 5)4$' ( .# ( ( ( ̻͌ǏՏ
<Ɖɐ
'O>K:C>7Q ˽ВңбΥϓՏ )M8IK:B?JQ
= ƙɭßɭ ƋĄȡȢ$ɭȣ-D.Bɭ :Ǎ ªÔ$ɭ ;
,âɭ¢ɭďɭ ƥ.ɭȤ-ɔB =
W w 5
6gj
6y:;$ɭ nɭǽï 6ɭɭÿ-.$ɭɚή ɭʼnȄή ƈùɭĉħɭ ɭn;\Go`
:Â
ªƚǾǓ6ɭ;
\h6 fɐ ҡϱՏ Đńɭ ÓIc
ĥНsӮՏ ǐñ
(NÎɐ bɭ Ţɇ ,ȴǝɭdɭ#,*/
5
b;[I
U` ,ȳǜɭɭĎš½ŵɭ
*ßɐ ѝmՏ U` \hqq
áɭI1Qw
C 4Ôɭ;
\gnn
£ɭ"+).
Ęɭ a=ZGq
dg
,QȞǞɭQɭ \hpn
$ 8h|
ÿƕTdtHOGm
6:. $ɭɭßɭ=ɕɊ ɭ ' ÝȽïŔŘEɐ 2
#
XU\W' 5
ɭ
¡2ɐ #`
Ö´Ùή Áɭ
KÜɮ 9gkI
ɭ µ
Eè s
5ň :ȟɮuPG
j\;cI
}EB ɭɭɭd:ȊnǺ ɭ$ɭɭ#ǟ
ɭ¡ɭɭ -ÿɭá
]
2
^%
AŃ J )I
ɭvPI
;@ ȱ d
ǥ
]
5Ń
$
Õ´ưή Ňɇ ɇ 5
"
ɐ
ɃƝɐ
E è *
"
5Ń&0
;Â
@ɋɭ Njɇ 'yɭ ;
,ăïĆɭ -ņɭ Êǹ ɭ Ǒɭ .iaɭ ҢϲՏ ,$$ga:yɭQiƛ¥ ǒQ$ɭ}RJc
Uo
ƺĆfbɭ'²ɭ ÁsŅɭ 5
"0
6
$1
(NÏɐ sŶɭ
Æ ɮ ƛ
oJɭ wɭ ɭ -
Jȃ Võɭ XyɭcyaAIl
hM
ZcI -fɐ ɐÍÖbè |ЙԇՏ ɭ=.
ȿɭ ɭ¥ ã gɭ .ia)ɭ¢ɭg ȵɭ $1\.d$ɭ
5
ŷɭ §Ǽ ̚ñՏՏ ɇ ïՏ ɇ
A è ƞɐ
Bè
+,ɮ w-ɭ ɭ 4ʼn
C âČ
żɐ Ʈļɇ nzD
ɭ $Bɭ:Տ ăÚɭ:Տ 2
ºÂ
Ǒɮ ɭ
Eè
, á
5)Ń
ɭïՏ v;c
:Տ =.$
)
KÝɮ Ɔt
Ylj
%1MǍŨǍǍ ƓNJUǍ.ǍǍ ǍŃǍŇǍ ǍǎՏ NǍŪ(Ǎq¸ǍưǍ%-Ǎ º ćMÆǍ 1fMǍ @ VXu DĘ%Ě-ÇǍ ij. Ē 1ĉǍ Jĵɐ
Č.ǍǍ Ǎ vǍ«RǍ 80© kÖՏ ƌ<Ǎſƀ%ěǍǍ
Â
7
TĮ
N ɮ
Ƹ
`ɭ ƥ¯kǍ%-1MǍǍ ¦Ǎ ;ҹԟжєÖԃ;ӌ΅Տ Ǵr ŲϾ&;Տ )ĞǍǍ ƦĽũĺǍ VǍ Ǎ àÿӈàђmՏ ¦²¾ƥ2ǚǍ · q ǍǍ ǂ$Ǎ Ǎ:EǍ º ğġÜǍ FǍ .Ǎ )ĠǍ tƕ2ǚĝ<ǍǍ Š&Ǎ /Ǎ Ǎ ĝrè;Տ Ā-ǍRǍň ¯[ĵǍǍ oɮ 2ǚ ͣՏ (ʼn3¸&LjÝ
Ǥɇ
͒АϑՏ ԮёԜϒՏ <ǍrɮƿoɮDžǗɮ (
6
$
ƨïØή Ɋĩ
ǐĭ
)
P ' Ç ɮ > Ǎ Ǒ
4
s
+,ɮ Tį
Zı
0¶» FmG» Y» ¤» ¥[I» 2vG» ·awveZ» n» 6G¦·GI» x » 2~B» y » 2¤» 2~Cz» §b{I 2C» I2:[» ·c| » R» ~I» [¯ » 1\2¨» d » ¦\I» :\2:G» ¦[2¦» ©\I¸» yHI¤! »
»
OÉɐ ]Ȇɐ
9 ƌƜɐ Ʉnj &e²I» 2» ¦e2YmI» tu"#$» nI¦»
ƱËɐ ]Ɉɐ
2B» ̹ĽՏ 6I» e~¥ » » ¥\F» cBI » )+¤ 2~D» )3` °:]» ¥^2¦» ą-Ǎ
=[
38 ¤ )J¤» șՏ 2~C» · 7I» ¥\I» :b:¯x:I~ªI » dI» ¦\F» :G~¦I» R» ¤[F» :j:°v ;e6FC» :e?nG» S»
ÊËMOý 2~B» wx)+3 ¤ I I:©e³Gn¹» -± I» · b~«G¡G:¦» od~I » )+¤ 5B» )3 G K<¦e´Ioº»
ēRǍ)+¤ > ɮ)3¤ ¬\I
5
# ɇ Glƙɮ
6
),¤ ɇ Glƚ
OÊɐ )+¤ Տ )3'¤
Íɐ )+¤ #ɇ 7Ƣń
Ĕvɭ F7jɮ
ɇ
Vhɐ +¤ 2B» ¸ĶƇ
Ɵɮ ɇ )4&
ſɇ
Fɇ
$%(+ *&( $"&+
/± I» 2» cZ\¦2~YnIC» ¦k2YmI» e¢» f~£:l8IE»e~» 2» :f:nL»R»2Ca± » » *I¦» (ɇ 2~C 2
YpF » 'U» ¥2» (ɇ Տ -¦3»ǹùɐ WB» ¦[I» 4I2» R» ¦[M» §g2~YmI»
% 1İ
+I¦» )/ 9
» ºш̴Տ
cÂ
ȎǂRȼńɐ
.N§» Ȟ »
NÂ
7Â
/%/&/,-.'/ %}C» ¦_I» µ2n¯O» RǍ øɇ
Dzɐ ţɐ º ,F¥
ȉƗdzƵȃɇ
ȋȌȍȎՏ u
Ǹɐ 9I»b¨ » 2=¯§I
?+ >* : Ɗɮ Oº º Aɐ " ' Aɐ
»
ÿɇ
8
9
Â
Ĭ
"ɮ2~C» 35`8I» HG~Ci:¯q2» PYwI~ » e~¥G I>¥hY» 2®» i¦» )ÞǍ /° I» ¦`2¥» 7LJɮՏ 7 4ɮ 2C» 1)Ǎ* » (V»2ms» ¦\I» i¨ » )¤+¤0 ©5` rfQ» ~» 2» @cAnI» X~C» §]G» rIY¦[» T» »́ǡ
=G[
71w
+ ĜǍ Ǎ#7Ó Ǎ Ǎ 0Տ ¬Ǎ Ʊ Ǎ°=>Ǎ Ƨ*Ǎ ǍŽ ǍƨǍ 8ºÈ7Ǎ ®0Ǎ , ūţ
] #Ǎ`'7 ~
BՏ +S9Č
~Ʉ~ɮBՏ Č
PǍ º BՏ
F ǍGĪƟǍ3!A Ǎ±Ǎ Wɐ p 9Č Č ǍƉǍǍśǍ !Ǎ Ǎ 6!Ǎ 0Տ "ǚ ƠǍ ɆǦՏ d*ńǍ
ǁ¢Ǎ Ǎ 'ǢՏ
) ĐTIǍ
%Čč
)4ǒɐǍ
%Č?
ǓɐǍǍǍ Ǎ Ǎƙ5ƍƄ
#Ô #^ǍČ G 9ÉǍ ŜPǍǍC Ǎ> Ǎ Č
eǍǍ+6Ǎ Ǎ0Ǎ,Ǎ Č WǍ +Č ,5 Ǎ Ǎ0$ǍǍ ǍUǍ Ǎ!Ǎp! ƚ (Ǎ2
»% ǍƔÀ#ÃǍ Ǎ šɇ $/TżǍǃ! Ǎp6ǍŚ5WǍ 3ǍšǍ(" ($/ CŰǍƁôÕ ƈǍn G Ǎ*ƲŊAǍ gɇ Ǎ$/ Ĕ Ǎ(/ F Č ǍnkuČ Տ "J ŝǍ Ǎ kǍ ¬ Ǎ Ch Ǎ Ǎn
CK
+ eŬǍ ǍW+AǍ ǍCǍ0Ǎı6ǍÁÊǍ9Ē i" $Ǎ b#Č` (8 đɇ Č ǍíXČ `Č Ì! bÍ 2 ;¾ǚ / j¦Ǎ óՏ Ǎ *Ǎ Ǎ Ǎ 6Ǎ¢ ǍA Ǎ+6Ǎ#Ǎ 7Ë A#Ǎ?
Ŕ\º]Ǎ Ē \#Ǎ 7 A 2Č čǍ ǍC ǍŕǍó ČɮǼ¸1Pɮ
  Â
AY CČ
ĊjĿ Ǎ ǍǍ `
ĶÌǍ Ǎ w=ǍƩ=ŹǍ;>Ǎ 4 ƪǍĨǍ¦t ǚź
}Ǎ
Ƈɮ§ƫß
đ|Ǎ +Ĵ$ǍŘAǍ/*Ǎ ǍǍ( Ʀwǚ!ƂǍ 3Ǎ Šɇ ĩGǍ Ǎ Č
ķý
;aǍ )¤
Տ
Ą {Č:ČÂĤɭ9Č)|ČĆ1Č d¨Ǎ Ǎ +lǍ Ǎ =3Ǎ 4(Ǎ )
$ /Տ Տ nƜǍ
Ǎ Տ &
%¤ )¤ ¦ɇ )¤
Ƭŀ Ǎ
YJ eŭƅPǍ mǍ /±Ǎ Ǎ =GǍ ,ƭǍ !(Ǎ Ǎ ƎƝƃǍ }Ǎ Ǎ rǍ h§Ǎ Ǎ ! Ǎ Ǎ vǍ WǍ 8÷ÖŮŅǍŻ$ƆÂàǍ \ęøǍ ģǍƼů4(Ǎ Ǎw* ǍǍ®ǍĻǍ ŋǍI0 Ǎ Ǎ á
ò˸ή Ǎ vǍ ơƞh¨PǍ ƇǍ Ǎ wľ iǍ GmãǍ ǝ_ɐ ŌǍ ³ǍƳ>Ǎ ļǍ TƴǍ TǍ 0ÍǍ Ǎ AǍ " _Wɐ ΏՏ ,ǍņōƵǍ Ǎ ŢCǍ5řƶǍ Ə âǍ
ǍƮþǍ
Č
Ē,Ǎ /i$Ǎ ,$Ǎ6Ǎ 0Ǎ ?ǍƕIJ´
[(.4` ǚ ý bðÂý ="_6ǚ ~Y© CÁý ¯$ý 8İɐ
?ý ý <Wý Ӥ>Տ = "ý )¤
=Džņ«6 ǚ "
ƞ(ǚ͝Տ ý ., ½ɇ 2Y©.F` Տ Gǎή ¬ýµý ó8ý +ǚƅÜǚ ȣ`6ǚȺɐf©aS© ®E/ý ,ǚ¶ý #Cý E"
]v© ZpPgx^
Fɇ
žɇ
Ćª«ɇ ņɇ
gɐ $(+ + `T© \_S© LSq~© NiOqW© *Hý ý *Jý Aªý Ŏe ǚ XiBº ŴƖɇ _6~ǚ ȈǠɇ ϥՏ U
vǚ L`©Dy^Vy© ~© 'ě8i^« ý ý Nj_Ǎ N Ǎ ÃTÈý O~ý ý _ŗ(ǚ © , © ,© 8ý ՆŹՏ "Ôý ##ý Ký ŷ
eD _ Sxeý ïý *Hqý ý "ûFý V ýý 'ý ~Y©0©
¡ q6ƻǚ ¥©£§© )+ +)+ + i·4ý +)++ &%+ `6ǚ 'ý = èý ^DDý Õý © Bñ"Éý Ĥ>Տ )+ + E yvMV© +TDr©siy© ~© ý b>
*)+/`¦£¨© > {pý
(º
+Îý bV© sV©Y
)+ ĵ <8ıɐ
X º @ɐ
Â
Û¢ æČ ! '#+ + @ČȤȅ [ ɐ *+ h© aȗgǤȡaǟǯǩɐ рՏ ý " ++DQj© |5zý +>¡ý Dý é(
Č
9
#( uÇǚǚ ( ( f(ǚ+%ƝCǚǚ¶DŽMƋ %¯ǚ ƧŬ¯ǚ +ǚ = l
Mǚ %$( (
ǚ ՏƔf»ɇ ĚųKǚ (ªǚ`ĺ(
( '( MǚC ºɇ
¹Â
R h S i j T U k V l m
W R X Y Z [
r n o i p st
=1' 9 6a ŕQ\ǚ ǚ Ɖ 62ǚ4Ɩǚǚ \·ǚ4^ǚƨ2ŴÂǚ 211wƕf¼ɇ szǚ ³ǚ \CMƟzƩǚ ŵǚƊŌǚ
(ůǚ ţ¸(ǚ4Ûǚ tōǚ ( ǚ Ŷƪǚ CM \%ǚǚ °"2ǚ4ǚ #ǚ Ņ2ǚ4ǚ
ǚ 4&Ɨǚ &( :vǚ !( aǏafCǚ ƺǚ4şǚ zǚ 4&¸ǚ LǚŖ Uðǚ
Ƣɇ
ǵ "
*ĒɮƀՏɇ ( ( *CĽǚ f ǚ !"
(ǚ ƴɇ (ǚ ªǚ ĺ?ŏƕ1 ƕ %ƕ
*Ƃɮ-#$ )ή!#% ( 16
Ţɐ
W5p Xɇ 5Wg"6"e W"oKe5 xCW25 4aX25 5"/S W17 Wpx S85 ;C {* .xq/ űɇ 5Wf"5"e
Y"oKf5 xo O5 kW22f5 {+ "p2 P5p 25e55 S5 kW22f5 {+ 5{5" UW 5| Dx 5&/V W25 xD P5 9eWpK {xf§Lxp Zo2 " 7 ¢Q55 / F R5 %5" xE :M_xr x-'`v;2 A 5{5"WpK SW }x.525 [pJsW5f¨ l"p© Wk5
L
{{x5
£P55 ɇ \ " ~xWW5 ]o5M5 "}Č
1
2Â
bNZb ,5 oxro<M"W5 ^o5M5 .O S" Wƪɇ Đɇ 'Տ +Dǚ 'Տ ɇ
Kº 6ɇ ɇ ɇ !&S øČ Bɇ Տ ¬F+Ń S5 (j5 xE 6 3
Dbå
5 Տ -5 " {zWW= Ww>K5 $* cpx¤o T" ¤O?o55 R5p õBČՏ !ɐ ʫʬՏ Տ Ex
Döä
yoW25 Wt5M5 ª :Č
4
5
6
# ՋՏ .DžՏ Wo2 T5 k#ef5 {xW-f5 ¡$f5 xF Dz
«
Â
VN])bJ Ⱦɐ EՏ ǣ!ɐ ĵՏ Dx&he !# ՊՏ
<ή {"W0f5 W WpWW#ee§¬"
mx5 x Űɇ "3b#/5p \u5N5
Wp S5 o5¦ 5{ !S$ $* O5 B5/52 pm-5 xG 5{ W ¥Wei #d5 x @"/S Hx S; I Wn5
əՏ
<3NJ,<:IS ? ' º &¯×ƕ E &è
'ƕ ]ƕ ®P¥ƕ úŢƕ L?==ºɇ ? L=ºɇ ? \X=º N<ƕ K?==º
ɇ _>º I MZKXº ¡8-#1´xƕ '*ƕ *ƕ C ƕƕ ? \X=º #KZNX 'ɇ L?==º I@\W.º
&Ƃ-Øƕ ("( uL=?Rº( = º Ĵƕ ±º}ƕ K=@R/º LC8 ƕ²º-Տ L=?SºƤɐ=0º '*ƕ # ³ ɐ ɇ ¹? 3 K ƕ ³T" ?º( ?º ƕ ¥ƕ ƕƕ ßƕ ƕ ²U³?º ɇ ǝv«ή ƕ#(( =º? ?4º º L?# ³º# L=?Rº ¤ƕ Ź»ƕ "(ɇ ?º ƕ ±ºɇ? , R' &h¯-şZƕ ƯÈĪ
5
]<ƕ Iuƕ ƕ ƕīċĻ ƕ °ƕŠřčůƕ Ƒ}ƕ$( '( %( -C -ČļTňƒàƕ ƫÇɐ ]ēƕ E Nƕƕ ƕ ^ ƕƕ =ƕ^oƕ a=·º (b=¸º E8#Ľź´È V 1 &ƃÙƕ 5 !
ƕ Ɠƕ8 1ƕ=9ƕR¼ L=?Rº? _=Rº ? \_?º-Տ @\\VºG K=?Rº @_=S A\_?º ? \\V º G
L=@Sº ? _=R ¬$ ? \_B? \\V ®º( I?<¬º \®º
J=CRº?_=R ºD\_? ºJ? \\V ºG L=CSº ? \_E @_=S ? \\V º ?
O=@S! ? \`B ³" ? _=R# ʑVÆ ԿՏ(" '
L ƕ L=?Rº F_>Sº"? \_?º# ? ]^Vºƕ <ľŻĄPƕ ąHƕ \º 4ɇ KcºՏ K=Rº ƕ1šoƕ Ś żěƕ-ƕ k¸ɇ W2 &!Úƕ vJ ¼
òĚƄ?Űƕűƕėŝù ƕ º `'( ~ƕ RÉ ±´PºJ' # '(ɇ µº ? ± ¶º)ή ?
K ƕRƕɇ =º *hƕ ƕEƕĪ P³ƕ Q Hƕ ŲĘƕ eBÊƕ L¬ś8 ƕ=Տ ƥɐ =º + ¦ƕ /N
pΓՏ Ŀ?fHƕûœHƕ ęƕ ʼn zƕ
'ƕ*ƕ ȴr''ý şɇ Ĕ1ŀũ ¡ƕ5ƕ S ƕ=Տ ¬Ďƕ*NƕՏ?º ˏՏ 5ƕ
Ǟ ɐ &'(J? º µº > R 5ƕ ƕĩ4Hƕoƕ ±4º ƕ #ƕ įՏJ?º ɇ =º [K ՀՏ Տ@5º M*ƕՏ(
&°[ƕ A
(
²{𠢺* /öğϮ̜Տ- ğÕğ *.
FXº
E
Gº
Hº
)7& Gº
ğ"¤ºόՏ zğύՏ
/´N - Öğ
=×eøCØ'ğ
&
Fºħ & ɍɇɐ
` ?ğ 2jğ ýċ+%â]ğ & ɇ j?GXº ¢º ğ\ğ = ğ ëe|Ëğ *kğ l¨ğ - }ğÙğ¼}bğ &&
|Ņƕ& & +
Ă'ğ : ă/B ğϷՏ ß ®º & `@FXº `º > * & zµğ Cğ & EɮD>ğ & i?FXº Ƅq
¨DŽ 3Í ]ğ O
³~ğ B 9"ğ +ğ \VZ]
ɡɮ ğ
& *.ğ"!&
FɄĩɐ G
*ÝBğ
& '
"ğ& & ~.¾ćğ lĄKğ & & °/Ďğ {.¿Ĉ
" & ɇ Zğ½ğ [C¢ÀPğ % &
#&
&4
_B-L
-.('"/ gԔԕ˨̼˶Տ
£"ğ&fr "ğ& &H $ "<' S Ûɮ ~ƕ ɢĨ
$&
£ õ 3 ©ğ Ö´Ùȅ ǣɇ
¦Տ
&
Ė Ǎ ŤǍ.Ǎ ?<Ǎ ǍƛĸǍǍ KéǍ ĤǍ ŎǍO5·Ǎ«nǍ ŏǍ4s²4 Ǎ5Ǎ œĻɮ ėµǍ ŀՏ > ̣͞Տ̸͢NjՏ ͋>Տ ©8c
`]Ǎ > ĥǍ
2gǍ
g 2fǍ
ĿՏ
ĬªǍ Ɠ ȵՏ aÂ
v
ɇ )2µ D9Ǎ
a j8cóaÄǍ Ɣ Oªõ8cǍ
K ]Ǎ ː
>
ò×JǍ
ɉɮ ĺ
ɁĦ ɀ Ⱦ
a6 ā©Ư|ùǍ +ãŇ Æ Ǎ Տ Ǎ Տ .V Ǎ¶Ǎ[+B³Ǎ?<Ǎ?[BǍ< ²Ǎ`ú99Xæ/çǍ Ǎ ǍŞB ǍǍ ŐǍ4ǍB
{ű! Ǎ Ǎ ,Ų+UpǍ X??&ÎǍ X 5{ }_Ǎ Fő 9Ǎ ʿՏ %ɇ ʾՏ Տ ʽՏ DǍ O.Ǎ Տ ͉Տ %ɇ - ¤Č Ⱥɇ
÷ɇ
Üɇ
͊ҞՏ ǍŁťV4Ǎ ¶Ǎ ~Ǎ+ŒƷǍųǍ
J9 Ă|Bû
>
ýɇ
%4
O îŎ
ΤՏ
˲Տ ÆϋԁՏ )ǍīǍ )¿ǍǍ ǍXBƐŸ4 Ǎ Ǎ ȗՏ ΈαՏ
f½Ǎ ƑǍ:EǍ Ɩ s{ $èǍ FǍ
@AL %)Ø %)Ǎ nj%E Ù ăǍ 8:-Ú %-ÅǍ Ć¾_ >
I
Lº
KǍ
,3è
>
aǍ
I
#;
#)* tT©>©JR©>©LT©cT©
oTNnx© đϰՏՏ GyR© ̷ļՏ oǓɇ &,©!$ &' )* dTy©?>©
kz398<©$©
Տ
MÂ
,L
$4
lx4::?©Ěɇ 3;;=©%©58 :>©
ħǚ Ê1_Ē1»Ǎ < 5&,©aĒ ̙ԀՏ [qu
© 8$9B Â`Ē Տ %1¼êǍ
'y¡T"© .
&ÁğC
,L [FÂğ$ ';
ěɇ & * =
1®ή $ B FĒ ğïÛąğ HT©
#;
*
(;
;
Î tÏĉãðH!ğ
; $; ); (TD©%© Šɐ ɇ ƻɮ ȳɇ ƺɮ$©/ ÿĄ ē&w%ǚ //G )&ğ
;,Տ <Ʌɮ dz¶æɮ Տ ! DŽ<ņɮ @cTy© Ǵ ·çɮ%© `Â
& & <ɮ 7; dzŷ%×ɐ ğȖՏ ŗՏ 8; FxR© \ŻȻÄɐ* Ňɮ @eT{
& 'x¢X#© Wǃ Dɇ
İɇ
¸äoÈiğ 6)%© ËĒ D|R© 7.©%©3*© %Þğ &"%* ·1Å&Ē DxP©
&Ē 1&Ē $ ęɇ 5A Ä&Ē ,=©
,
ˣ̀Տ
-=©
/*<©DT©iwiqD© AcTy
( ;
&ɇ +<©$B© ή$©Z7º
* oĆ3þªğ 0 @cT© !* hՏ ǍcT© (* f?r <ğ L¤©cT©qmyT
Տ
C$©: ; ıՏ Տ %& p ušǔɐ ɇ :
%.;
MƕĀ§ƕ##ƕBhcw=YeJF
Wc
«Տ ƕWÖý ¨Bƕ¼É½Í>bÈ^X;Í î#Ĝvƕ Twn
m;ET{r
Xs
ԍ÷ϜՏ FVrw;fCI
Nmh`
Џý BţŁ4ƕ XÍZ Í ¨Bƕ +-b ʽÍɭ^ZÍ
ZÍ ` Տ (7 Í ^WklÍ
Dzɇ K
Í÷ϝՏ Βo×Տ ҦϳՏ -ƕŊ ƕ ď4#ƕ4ƕĵƕ ([)b ɭ !' 7 > Zj<Í ȹɇ
^= &±9[ƕ ^X> "ƕ 3<ɇ Տ
ȷɐ èՏ 3õɇՏ
b
ǀëǍ F
(Տ IWb bJ*-b ?K` bLab.MXL) K'b b£l
S(Í>C m'Í.NPb>bxzÍb _kÍ ŖՏ
>b ^Í1 :SLc
6Տ Տ ';;cF
Ūƕ¤¨Íɇ ^Y?Í
7Ǻ
^@ 5es~Im-
^Z^D äüƕ^Í ɇ ɇ ^ZZXAÍ а hý uՏ > NPRb
ah ¥¨Í
¦¨ Í ¾X^Í©!¨©
7>nK
kÍ ^[\ZÍ ʻʼՏ ɐ ɐ kXZZ=Í ôj4ƕ ɇ K ^¦©®BÍ ¢²°³ Nb P S>b:b
8
AO\Ub.=b kXZXÍ ʹʺՏ ɐƗƘɐ `X^. M c
Șɭɭ kZXC
9ĝ%Bj½ƕ §ƕāĞƕ
3
667:^
NPTb¡Í ¿Z^Í
4 =% bČ
Í Í Sº ^ Í :bb.b Í :b Í
Տ ÍÍzÍÍ^Í F ^Z^Í ĐƕĆğË
m
§ K w 8 x # )&𠧺ğ m L3
j g"1 g@
£ǐɇ
<ďé ğ Ʋɇ ğ§=º NSǍ ,ğ®ğ Lğ^?:Nm
g0 Y g@2 i@"J" §º Y? Y Pb qS Y Pb $𠤤Ǎ
ğ
6:=Uf&'m m )Ð&ğ Vfm ğÃ(ğ ğÇğ#ğ ,?9M
ğ
Ƹ¤ɇ
ğěğ 1L6m ğ(m
Ƴɇ EՏ mɇb ğ P
9ğT5ğ ?:ğ ğlm -@`\m
üJ ğ ğ/,0 ŔՏ ( ? Ĝğğ#4ğ #ğ
+GK4:mu % Àɐ ğğ 0O7m u ğ ğm ԲmՏ ?/a:mû ğ ğG(ğ ğČ ğ Vgm ğğ ğ6ğ .?:m I1X=:]^m vsYYD@Ɠɇ
IžǍ Ñğ Um ğVT'ğ ğm ğU7ğ $m *Od ] :W%m Qğ
[A2 YB Z$@ĉ ǺɆɐY Y@Z Ƣɐ Ǚɐ
)Fğ mğğ
,?8P
»jğ Ėğ w
;Ağ w |Đğ 0!vp0£Ǎ ´ƌ {4ǚ ´+ǚ f m ğ ì(ğN ğ ğm = ğ ğ ğ)m L1J:Iim , ğ ¯đx0@ėğ ŴSǍ , ΚŴՏ ğ ?8 Ǎ Ċğ å ğ T3bET`Ym _?1_m / ƶɐ
ºàpğ W7 ğ ɇ ,Xğ` Ɍɐ ğWcXğ ntç ğ
1 ' L w Ċɇ
Y
ʷʸՏ VR
ZA2 YA#5
Ǎjm ğ hm -A:OmÒ¶ ğ ğJğ $ ğ+òğG!ğ ,BFZm
]CQe[m Ǎğ ğ 1m ) &ğ S[3*ƒüǍ/£$
ĕ ğ ğ#ğ
jm ğ
Gğ Տ /ÁĊ ğ ğ+( #ğ ğU! `gğ 7;ŕՏıՏ *Ǎ *0Ĺ$ÏǍ 3ğ n`E _m ĭƽǍ ę ğ 9ğğ $ÿğ 1IIjm _ğ · ğJ ğ ğ
z q {Â
ğ ğ ğ6ğ
Eēy«ğ ¦P
Ǎ 4ğ Ęğ Å÷<ğ Iğ :lj 8m Ǎ+m°Ǎ$ğá @Ä2ğ ğ 2gîğ
1O6m H5ğ Ā v2>ñHğ . ß
ğ ğőħ ɐՏ Q8Ěğ ğ76 ^ ğ ;25?mmğğMğm mm s æÌyğ "jm ğ8Iğ ğí5ğ èO!4ğ !jmJ8Iğ ğHm Ħ m ĮՏ
Ø'Ē
Č
!m ɀ·ɇ
Aή įՏ ɇ ɓӠՏ 5 ή Ɂɇ ɇ ȩĨɐ ő$ή -&5?ͯή ˎՏ -%, ? Ǻáǀή ȁɇ ,Տ 6? ɇ ,? ͑4A $ή ȇɇ ɇ Ġή ǵǶήǁή ˽˻IJɫǷή ɻ 'ήŘή .,#? )ήή Hƚ ή Ͱή > ή Hm ~~H' hή
SA HÅή $Ç8ÎήXή
Hg ή ή
Ñɇ
".,$?b Ưή Տ L? ήb??? bbb6b b/ǚ bGή Տ Gáή
bAĹɇ
ǻ <$ 'Ńή Z .LJ A>M ή ')ɇ ') X'ή~dή k¹ɇ ðćή ӥđќϐՏ Hm ķɇ Υǂή .ŸՏ ȦƝǃή (4'ήƖˮ .ή {$z(ή ) ) ) ЯˀˁՏ ,
>z4ή (ή ) ɇ ! ) ") ɇ Ĭɇ $ ή #) Տ $) H$ ή ') ɐ ɐ ]Xή'))%)$ )#) Aή SA .HÅ /ή'_>' ή/$*ų$ ή
x ºw
}ɇ ğb
<$ 'Ʉή ȹŁį
4 JK
Ȳɇ ķή ɇ Z.Lj
}ɇ
Mɐ '$Idήz$ή8ήX'^ ή> $Iήğήľğή ɐ ͊ήGήmŦk.){
Aή4ή.Hή $>8͋ή ή
I )ή6b H )ή Pɇ
ɑՏ ńɇ ğ ɇ
y!)ή
ɇ ɔZYdžՏ
Ǽ <$ Ò'Ʌή GGĮ Ō(ή = ˆˇՏ ή 8 ήʍ$ήÉ(-'DŽή ̙ ή~ͱή ͈ċՏ cÉήή5 Ś'ήŢή ήEή$ή Տ ͜ή6ŶՏǂǀǁՏ ŵՏ6ċՏ H
z'{8 Džή(0kƕƕ˯.ˠή 6Տ ƆH$ ή ͌ Ǹή ï$ 'ήɼ̚Ƶȧή Ŧ͍ή ') ˄˅Տ!
ɡ ή ë ή ) ή ̶HúÉ ή Ï İή ¹ũc$ή(ū͎ή ʎ'ή ɒՏ zĄ˰ʥ ή ˇĄ'ή >Ãɇ Œ džή 4ή (ŗÅή ÉĐř͏ή Xή ʏ ˈ¢"ή4{ ήśŗ ή ή °ơ̆h ï ή!? ) ˱ ή5ήf$ ή' ǹή A˲$øʼή.4ήz̭¢>'ήĄŢή>ɇ ú ή>$˳_FÄƙή ͐˴$VήǠɐ kƕøFÏήX Ā ɇ
ū{ ή ή_>΅$ή¢ή4ήzʐ ή 4$ή4@͑ήʑÏή Ö̜ήƕ ; ®ή ¹4ήx) ̝>8'ήÀή Ρ ήÀ͒
4 ήøʒ ή ή FZ X_ ɐ
1>o 9UP:Nk;?o ģЎÐģՏ ªѹɐՏ Տ ªѺ ĄՏ Տ I
Ðü
:ɮȈȳɮ ßѐîΑՏ gE2fo
}ý ;
êý ɇ 7(§ !¢ý
2Sο ή$>̊¿ÎήUBoTHg͝ή ü y Տ /b ȱɇ oՏ 78b
Ký Տ WŷՏ 0ӢՏ Տ o
XÔÂ ˢѼyԱϊĚʤ ų ī ɐ Dɇ
ąɇ
Ņɇ
öăɇ
(WXOnogG@odTή ̮ˉήΣқӟЮѻωՏ TPo Ιƿ͜ µ͡Տ ΆΰՏ RNý _>dYA7hIl>Nn% Zo ɇ ¥ɏՏ ɐ 0% Z¿ Ĉ ! Z ] o ɇ o *Տ 1% ^ +30% ] 0F>PoZo Տ o3Q=o ]o Տ 7. '§ '[\NnJRDo iE>o HTήTCo8UeKP>oLPo Rý͛ՏͨՏ 4D4MPoή V6f5R
² )Pbm>a&o 5% +?io -o8 ¿ ήI-'S E%0PS ( *c o /H>P / W%b - o^Ǎ/8%)b > "oîɐE )b Ĝɇ -o Տ ;%!
,>jo ǰɇ 8@ήŪ¿ή ĚÐæüԛyՏҜϭ jƆŅɮ NjǾǰȞ ^Ǎ è
Ĭ 29% 5 9œՏ
#[) $ ` o ɇ !o ĝɇ DZɇ Տ 4%"
2,b
³# :ńή % h A}ź*ή ƛή ĕή)ήƛή Q Ɲή > ; Û ήQ
QՏ
>
ɐ /+ή #ÿή I+ή Ńɐ ¸ ƛՏՈՏ ȹɐ
ʄή
ɐ
Ɉɉ I˨ʀ` ɵ Ɋɋā Ɍɍ ; ; ɒ ή .Â
?
:zɮĔQƜή)ήĔQȢΧή ƙɐ Ĝή 5ήQΨή
ĩή hή ȫ Ý ; ɩ˗W# ɶ Ɏɏā ɐɑZǪή
?
Σ Q Ɯή ήOή Α˘oŜήW#ήŴÓήo/ήΗ `ή ήű.ʴI7ήy TWWή Ǖ
&ή âή Aή (9ή ήb:ή /ή# D#ή ƀή ΤQĕή)ήĔQȣ ĕή > ή D#ή ; ή u ?> ; aȸĩ <` +ɀή ȥľǫ ή)ή ȴɐ / ȳɐ ßή ¸¡ή CL Ňή ?
ήĖήìή µ ή Ň ή aή aή
>
Ǭή
AƏή èȝή ήȞή)ήΦȟ έή ¹ % #/ )ή1 ɺ̑ή © )ή & #/ )ή* ƪ©ʉȠήęήĖȡ ×ή
]Â
_Â
% /
͇Տ% ,/
^Â
·ă ή ή !Ōřɐ ɇ ; Ī ή ¸ήÀ"ή Ñή#D oÑήo¡ \:ή üɐ }ǟũƱ
>
%/ Ęή }ĥŗŪɮƲɮ* Ì҉ Ɯ ; ǯ
Aomή ǫ̆ ìή ɐ ʙ ή Əbɮ Տ S ºƕWήlήÊ+
ή ̩ ή¨Ä> ή ή èȳ
>
ai ; â ή
ĺ <̐` W"Ņή Kή ͂ήo#ή ;ή |ɇ
#'
4ɇ
Ȥή ɇ # Ωή íή à
aȺ <¡`ΒƉêή Jâ ɮ ή ήI ήVή ή\ ή/ή CWήbΘήřήŖήƜbɭЌ Ų Տ :ʵ
tΞǭή ¶ ή ήŖ )ή'?ЋՏ @ K9Wή CWήtWή /ή ήbΙή˧`
Ǐɮ ũɐ ` iή
@
ɠɮ
4
4 ɟɮ?
ɋɐ ĺ ȍή ? ? ? /Â
Â
ďɇ
5ɮ
üɇ íή
¸ήWĆĐÄo Ƙήŵ `ή]/ή ňŪɐìή U` þɐ !Ǯ Ĝή ʊ ή1 ή ͻ)γՏ БՏ Տ ƂʼnUŅɐ Ľή
ɞ
»
ɇ
ȬήíήȰ ή
»
VÂ
N e O d e P O f g Q
ī &>ɝ2ťɭ _Ĭ @ɭ Ýɭ ȥCê+>ɭ¡ɭ ɭ ǭ ɭ Ië>, ƻɭ'
ȅɮ ò#ɭ TȦúɭɭ p J ĭ ɭ Ɖ2ɭ öЭՏ ɇ
0,
K%ɭv+ɧɭ 2P+Gĉ¶ɭ
M 2ɭɭ 2 ɭ ö ʵʶՏ ȧúɭĒ
ɮƕ
2.7ƕ
ɣɮՏ Ɓ ƖƗɮ Čƕ ƹɮt Ŕ h t ɭŕʼn Ùó
ЬՏ tɭŊěɭ0 · ɭɭ_Į
Mæɭ =+<¤%ɭ :b>ɭ Gɭ ȇɮɭ _qİ ÈR?Ŧɭ L ñ
ÍɭH>ɭ Ú:2)ɭ Ǧɭ | Ĉ | ¶ çë ` & Č| ɪ |ɭ ʳʴՏ Rğ Îû2ɭ
*eɮ[
ɭźɭƎɭ Ē
ĒŐɇ
ɋe
`uɐ
=
`u uruIɮ
Տ ŏɇIɮq
ǮY.ɭ ǿ«>>ɭ dɭǡ2Z);ǯ%¸ɭ ơ>ĄóɭNɭ ɇ ʱʲՏ iáǚ ƌ ɭ Տ tɭ o p ɭ Â ĮՏ ɦɧՏ įɭ
ƊìƦɭ
_Śɭ
°
Ǖ¾ Ħɐ
!% / / ·æՏ >fh(:Ȩ%
!ɭN/ɭ ^ pŇɭӞՏ ɇ ӝՏ K
ý ý ý Dɭ
5 <
& ą &½ɐ
6< ?
?
<
9<
T & H&K^ &^ &ÐŤ V <:< !< < Ć &ɐ &,IՏ
(K
<@Ýý ý ɇ ¸SÞ(ý !$ý òG¥IąIɭ hɇ Տ ²IJZήOɭ«'OƱɭ ý :< Տ & ý 6< "<
LÂ
ŏňijɭ
,' &R!)ŧɭ /6 7=K £ý ɭ ȫɭ :ý ɭ ) íɭ û' Dɭ 0`ý Nɭ jNɭ & (ý 0>K / 7= 7=K #< ƃɮ Ɛɮ 7BK :ßý ]ý 1 < & !Ñ Qý
7?K gҧ IՏ Տ 7CK ý jɭ < ký
!;ý
7BK
< '28K 1D
4 4 & f ?7?8? ? K4k 7BK
U< < ;< ģɐ v ^Í
óɇ
Տ a²ʂή
Â
7=K Rɭ ìAý ͆Տ !< (3D
Â
u
Uuć 4v hX/
Տ /EK 7ý 7)K ɇ /6
&R!)Ũɭ 4'1 Q
7<
!<
"K M (", K
1< 61 :ɇ< Տ & 3 :ɇ<<;* !K < ' ǧՏ
2.< 2/<
<&4<ɐ 3 ";?)/ - 7ɐ Տ 7ɐ 3 < 2 " K ;@ ɠ8j º & < & 2< & && ğ Ĥ ĥ ɐ &
Â
7Aý 0ɇ
\Â
ɭ ý RIRɭ Ƽȩɭ + Nɭ Yɭ 18< Yý Xɇ 0` lý Z ( G m ý +". K º 2 < Øý P+O/ƽ}%
4" $K Ƶɐ 2< 9¤Sý & OɭYƾɭ \à ý &Z_ý Dí/O«mɭ " (",
\
< 0%$% ÑȋWɭ ƈɐ
Uı &!Ȫũɭ ʁƽ 6íý Ş ɇ
M < "4 Kj
KÂ
4(/K
"K "5K ɇ 4"K
&*< 'ɭâɡDIJɭ Ñɭ#&ɐ6 < !<&6< Տ (6<-Տ Ž¶ɇ
!;Ò
# ! & , % " + < - ! ' ) *< 7Ågý ý ÷Ìý ɭ ɭ
<
; & #< $ & &
²ή
'ĢǍJìǍ '' L+ s- 5 ?. 3pu )ή n t. 6 ; $ # %' ' ' qr ' ~ ºLº2;'; ɲ̯:ƃΆ͓Üή ' $ 9̞·ƈή '; užʭή ! ' ';ųή ʦC˵§ ή ή OBC@H'S QSS ʯʰՏ ;u R+ή :
-'=
Տ
ʭʮՏ
ÀÂ
ŐՏ
Տ
S
E
JÂ
<ſ :éή L9 (u µʽ.ή 3S &( rel="nofollow">@J'S K.'S 'RC'#J'%S ѸԙћΘχĘՏ @*S FL'CFS |ή L
-'=S 47S Տ
SՏ PÄý
NS 2 H S$:'"S ͲC.ή S E
)FS M@S ,@S +@<S ήƙ
' * '&'
īЉЪЊՏ 1;D82'GS 59S Տ 6S
Ē /'?S Pý Ǎ ý $ƄɮE J ě ήǬ ǭ Ǯ Տþ $Ť ɮ E 4J ̐Տ ɇ Pɐ )ήS Տ
Տ S E
ɴ¦Տ
ή
'ɭ
' '
;u Ɛƃή LJή ήɇ Lºǩ Ǫ ǫ Տu;23 %u
@
Տ 9JuǍL5 )u
#$* ' &* ! ')* 0:'A?F"S K*"5M.3S 3Q6D1 S S BŒɐ %*,; 2($; (6%; Q 36,; Q 8$; Q
Q
vȷȷ0&ȷ ^˳˾źՏ ^Àȷ ˗Տ TMȷ*ȷ *±#êȷFȷȷ&â"#ȷ T]ȷȷ #ɮ &Ă0 1`"#ȷ Àȷ **ȷ Ƨ*ȷ ÓrNj?/ȷ vɇ ȷ /ȷ 0?ȷ TÁȷ ȷ % D/ȷ &* 9ªǦ0ȷÊȷ"ȷ0ȷ @ȷ `˻Տ ȷ ̃Տ 1*1ȷ Ȣɇ ȷ Ëȷ "$1ȷFnjȷ % ȷřËȷ B ]Śºȷ ȡ*ȷȷ T] ļȷÁÅ
Q ʼn"ȷ&&ȷ§ZȷFȷ01ȷ ¬ɮe ɇ ä ȷ
Q É#ȷ ţ û ɮ d ú
¶
·
¸
Â
V/ȷȷƑǯr* /ȷFȷQA16ȷ*&ȷ¤r'7#ǽ1ȷ¬ȷȷƖêȷȷ iɮ
òɇ ïȒɮiɮ
'
&'
fɮ ŏՏ £ ŎՏ » ĸ
Â
Տ
HdȷȐȷ *ȷȷȷJdL6#ȷ*0#ȷ ̧Տ ȷȷ 8 Տ Ųɮ ~ ѮˋъՏ ȭՏ ʀƻ
 µÂ
Տ 8ѯՏ
Â
ȲȳՏ
Â
vȷ ȷ Ɨ&&d0ȷ ¾ȷ Ȱɇ b ó ¶ȷ #ȷ ȷ de/ȷǘĂȷ ȷ ȯɇ ăՏ șV Mȷ ƹȷ ɮ Ďɇ Ȋɐ ȣɐ å"ȷ""ȷ MՏ ȷ &&ȷâ1ȷ# »ȷ v¬ȷȷ Jøų&ȷȑȷJ#ƘAZȷ*ȷ ¦#dȷȷƟȷV* Fȷƺ*øȷȆȷ9Ȓȷ2nq£ (ȷ/ç#ȷȷ#ȷǾóȮȷǿ/ȷr&§&ȷFȷ ˛
ů
LՏ
%ȷ
Տ
Տ
Տ
Տ
Տ
'
LՏ
SՏ
LՏ
Տ
ȔՏ
MՏ
Տ
aɮ 0ɮ
Տ ʇƼՏ
Â
ɝ f
Տ
£
MՏ
£
Տ
Տ
Տ
SՏ
£
Տ
Տ
Տ
Տ
Տ
Տ
ȷŭȷ¤0ȷFȷƙǧrªȷŻǗõ2û ȠƐ É#(ȷ Ե ŭ ՏԽՏ 7ȷƃöúȷh*y&ȷ5ǍVhȷżȷȓ(ȷ ȷȷ /0ȷFȷ ǂ&ȷJǩ1Ȣȷ (/ƥôȷ LJ´ɇ Je`6ȷ ¤/!#ȇȷ jµɇ Ǩ6ȷ(ȷ
ɵ©Տ
D¬V¾µÅtë 7TÑu¥TÐfTë E¶guÒßë -:%A
Q
Œúȷ
ȷ 9:azKfa f? ] ȷaNÚ(ȷ ȡɇ (ȷ 4ǒՏ HáIŽȷ $%'( Տ Ǻȣ ċɮ ÈՏ4Տ Lz , 7L,]::u f@ v ³ ÍûKȷ ̠È#ÊՏ ǻȤɮ zfG, 4˥Տ MzX ;[;?T:T
@ȷ!kļɮ x:1ȷ , ȷ ÂŗTÌȷ `ȷ ,b Mzfz4:T:z (ª,XȳW"Įȷ ̟υӾՏ fՏ 0:
Nȷ@Ǫ(ȷ fAȔNȷ n:rn:c5M2U-s CWȷ ÊՏ WǙ(@ȷ !kĽɮ UNIȷ `¿ ˌՏ Ņ] ͷѵήՏ #̓ÂbՏ N{ , Z(ů0?įȷ ŝƻǚžÙȷ ̡#̆4Տ ɇ ( Oz $Iȷ B,3 , zq,t9 ȀǞɇ ùÑ(ȷ #bՏ B b¿ ŞCWYNėȷÂ]ȷ Ɨ º·Տ ɮ!9 Ī ` À ˍՏ ̇Â Տ #º Ǔ -Â
Î<c ɮɮ/ɮ ɮɮ ƒ Տ ,c6G:c ɮ "Տ ăɮ@ɮՏ S xƨ1ȷ Zȷ(ÙȷȪWȷ f0NfzzfWOgaz µČɮ åɮ "Տ ³_Ąɮīß Vɮ´/ɮàľɮ Îą?ȷ1Wȷ(Ñ(ȷ
NȀÚȷ ,Y:cfzfV
Kfdz ȷ ɮƛɮĿ
ɍƹՏ
)onfz:@ɮ¬Տ ŀɮ #?@ɮ ±Տ /ɮɮ Տ ąɮ(è
ÍȚ®ȷ +Տ ɮ ɇ ɮ WCȷ ɮ Տ 0øɮ (£ȷ
9
ƕ Տ ΉεՏ GKz Kz cf ,ȷ zfTOfa
ЈШΩՏ K| Ƽ¢X@®®`Ò?Nİȷ
Ǝ
#C eɇ [Տ _ ,è"ȷ
˔Տ
V
ɲƷՏ
Ŋ'ȷÄ5- ȷęHźǼȲȷ útςҶĨÎюû{ԼūՏ ȷ6ȷ
ȷ Ě
ġɇ %Gȷ \ ̦Տ' % Ĵ%G
S
ȷƽɐ Ůůɐ %½kĩ ɿƵ
xȷ4ȷOȷǔĪȷ m ȷ ȷ4ư!PȷǺǻ ȷ7ȷ #¡ń Äc÷n4ȷ ȷǀp:!ȷñ4
Ŕ
%ȷ \ = ȷ ě % ȷ ȷ ȷ ȷ =ȷ %ȷ \ % ȷ ȷ \ % ȷ ȷ %ȷ %ȷ ȷ \%ȷ Ĝ=ȷ %ȷ =ȷ ȷ Ƒ Ȯ Տ Ɩ Տ ȷՏ ȷ \ % ȷ ȷ ĝ = ȷ =ȷ %ȷ ȷ ȷ Ğ ȷ = ȷ Տ
œi f ȷ ȷ 44Jn£!ȷ ' ȷ Ɣȷ ȷ Ȩȷ ò4 ÷ȷ ynȷ -'X5 ȷ ȷ 5ljȷ 'e5pȷ G ī ȷ Ŏ Ąȷ ȷ ȷȟ4ơȷ ȷ ȷ 5NJȷƱȷ ƲǕȤ© ȷ ԘtέσėՏ ȷ!P6ȷ X4 PĬȷ HƳcŹȷ ȷ eYƴ!Pȷ5'ȷPȷ ĶĒȷ ȷ-cȷ6ðȷÒ cȷūȷy©ò ȷ4ȷȷ ƍȷ ð:ȷy©ȷ'i? JƎȷȷG t ȷ ǜ Տ Ŗe ȷL ȷ
Sȷ Dȷ ġҖՏ 8 ēȷ ğȯՏ ȷ =ȷ Ðǖnȷ 4ȷ /ɇ ķĸȷ ·ȷ ğĉŋ RՏ S ÚġѴōэՏ +Տ άĉՏ 2Տ IJĔȷ P 8Տ 8 ġ « ȭɇ +Տ Ùɮ nȷ 9Տ Տ Ƀķɮ D£ȷP ȷiƝ-ƏȄȷ ȷȩȷ {ІÎ{Տ ÚŪՏÞՏ eՏ ͧƸՏ
m ȷ ӻѰՏ
7
Çȷ9Տ ˑՏ ijĕȷȏcȷƵ ñȷ Ȯɇ 7 ȷ җėՏՁՏ2Տ Sȷ ŜÐ ȷ ԻՏ "Տ ĭȷ DcȷӼѱՏ C 1) % 6āûĞЅՏ zɇ ƜɮȓƶՏ ͐ЇτtՏ }Տ "Տ eՏ tȷ ŋ'ȷĠĆ Ōь 6 }ĆՏ &Տ ŢĖȷ ƶ ȷǤǁǥāȷ ȷ zɇ eՏ ͦƴՏ ͏Ĩȷ ӽѲՏ &Տ iՏ ƕe4ȷ ȷ i²ɇ Dȷ -Ŭȅȷ }Տ "Տ ȷƷȷƸ'O4Sȷ
ɱʡՏ
šȷ ȷ ȷ ȷ uɮƣ Sȷ ő ȷ ĠѳՏ %Տ pɮ ȷ iɇ 6 dɇ dɇ ²ȷ% j ȷ,ɇ S ,ɇ S G t ȷ ̯ҔӸπՏ
#* 6 13 )6 6.5 & .+.) 46 Uȷ =T >U Sa %Տ iƱՏ UFȷ ƀɇ ɇ %Տ ¸ȷ uȷ ű|Yƅď Ā#ȷ B;\?a >a &Տ iƲՏ ōÜȷ7 ȷӹљŊы ŧՏ/' 6 hȷ Ćɮ oqȷPƣȷ >ȷ #^/ 2µȷ ȷ2ŧ ȷȷÖȬ ȷ ӷѭՏ 8 ȷ ȷ?&ȷ /ɇ ȷ 2 Ȍ.2 ǑĦȷ Uȷ qȷ ȷĞЄìњՏ «ȷ æȱhħȷ ŕ ȷ sȷ ` ȷ ĂՏ 6 ąՏ ʼnՏ ' Bȷ ,ɇ S TŽՏ 2/ýoǵȷƭ«ȷ Džǜdžȿƒǯǝɇ oȷ°/ȷ
ň¨ȷ 9ȷ hqMȷ ȷ Ǣȷ ΗԺՏ ùĈäԗΠӺЧҕĈՏ ȷ ȴɮ 2æȷ 7Õȷ Cfo .ȷȷ .+66$, ! ĀBì Ƒɇ @a* 6ªǒØɮ ЦӛՏ ȷf ȷ ȷ ƆƤȷ2ƛ# ·ȷ äρϼՏ$6- 66 /ɇ Տ .ȷ+ȷ >ȷ ŸƜb ȷÜȷ2 - 6 6 Շ / ǟՏ H 2Ƈȷ & Տȷ ùӜՏ ȷǶ ȷȷ oȷǮ ; ȷ $( 6 60' 6! 6p Ł
ƅ
/7)=:>$J D*$6D/2J 9,$DHJ ,8';9?$J D+$5D/0J 2I3=.#J J ş Ŏ Ŗ Ų ɐ =$7J $E.:7J />ADJ >9G7!J
6@>\@l>7} ʆՏ &z\A
Ehm
$\nriz9rH_\l r_ :_\rBor7\ro
ȫ ƭ Տ 5<0B ++Æ Æ ):59% 6B
ɋƮ ,9B ?$:+B 7=,5B B 9B 7<-B89B&4;B !B 5 B 9B '(0%(0 9B :7B %= ?$:-B "5<.7
ɰ Ư %B 79'5B /B "B 9%B:59@B A%:0B #7>05 B *:59%#B +27B OՏ јʹӄуư
%B : 9$17B 3B %=
( ( ( ( ̵͗̏͂Տ ͖̮͎̗̞Տ ( ˪Տ ( ( ( (
ɾȬՏ
Çƕ Mƕ Fŋƕ
ƕ ?º ɇ ɭ ɇ ɭ
) ƕ^ ƕƌC$ƕ ƕ
ƕ
Gº ɇ ɮ ɇ bǍ
<
Îƕ
ɭ Pɇ Տ ɇ
³Â ´Â
ņɐ
?º ƕ ağ Ïƕ ağ <
4 ƕĬ÷:ƕ3ƕƕ F3Ō8ņƕ%Ãƕ
ğ ɇ ğ5
ƕ ĉS£ƕՏ % æE ƕƕ ƐQƕm6lƕ@%ƕ wƕ ¹ƕ !aƕğ ƕAŷƕ66ƕ7ƕŎ(ŧx 1ô ğX _ » ɭ "ƕ ĥΎՏ'ɮ ;ƕgƕ,ɐ _ƕ!Ŗƕ 0ùóČ%Dğ %ƕƕ /l ƕ 6 8 Ē ŖĈɭĨ ɭՏ yƕ ë4 <ƕ ƕ ž7©ƕ
| CȁȜų J'ɮ`©¼ɐē ɮD J'ɮğr ·ƕ#GIƕ' - rɮ tƕ Ĕ ɮh1( Æğğ, »ɐ| Ĺ qðAÖƕ ȋɐ $ AGƕƕ _FŗŇªƕT,ƕSƕ ° ¹ɐ UĴ
LjȺ
ȡCǮɮ ¦ ƕ V< ɇcY Kğŝû ɇ ā ɇ ¦Fƕ ůɇ 2Q:ƕ COƕ Ɓ.aƕ Øɇ ɇ ƕ $ƕ qȲɮ Տ ƀ = Ŷ Ĺ "ƕ h҈ՏLɐ ǜ =ɐ DՏ qŴɮ `ȕɐ zȥȱɮǢǣɮ. Oƕ 0ɇ ɤɥՏ Ðyƕ ì. $ƕbĕƕ ƕ 0ɇ ª>ƕ %k:%ğ ".8 ƕİ _Ïý Ÿ 7Äƕ Â
 Â
  Â
6
= Iƍ tƕ Տ Òƕ ƕ ƕ
5ĤIƕ ¥ƾ
Ńɇ
ɇ Ȭɇ ɇ ëɇ
<ƕ ÓÜ
ºÜɐ Ǎ q ¤ ğ Úqğ 2ƕ :?čiğ ƕ 5ƕ C «h>ƕ bö
ǍAI[
"ƕO q Ǝuƕ(ƕƕjķ:fUƕgĖFĸőOƕĥ0 )Ŭ ƕ$Ĩ$ƕ`ƕ ƕğ ɢɣՏ Åƕ n ǚ ¦GƕńS! ƕıƕ>ƕgĭ $ğ Ȩɐ 0ǘ$Ȅ1 ǻ3 Ȓ ê0ğ dğ
ɇ tɇ ɇ
7ǻՏ
.$ƕ ɇ ĕ 5
- óƕ ÞƕĂſƕ>ƀƕ ƕ Mu 2ɇ ʠՏ E: ƕ ;:ā
±Ĕğ ³ƕ/Ũƕ øŞƕIJEƕ3 ƕƕ _@1ƕĊ =6 ƕ($á
/ "ƕ Ŝɇ B*ɇ ;ƕDŘTƕ7ƕ 0Qă£ƕG)bƕ>ŭƕNbğdğ < żɇՏ
-ɮÊú1YğÕp À ȶ¦ U³ɭȷg ŋɫ »ɭ çŮ3 ƕƕ®fƕA Uɭ
ƕ!
Hâ
Lý ý ЫՏ Ķɇ ǘ Ó ǚ Ǚ ǚ ý ý ХՏ Տ Ô ǚ@Õǚ
h ý öÚ
/Ăi.ãǚ Å Ŭý Xý ý 3 [òý 9ý ¨ý .
ä ǚ ƼǪ$ɐ ý ,ý ý # ýVý ý rùá ýý 2ý .ý
( ?ý Տ ÉǚHåǚ [ý +TĆU æǚ
@çǚ * ý « ý ô Fý ˱˼Տ ý ýBý h¹±ýMý Տ Q=º¥ý © ý ίӅīѷՏ ^ ¸ tý L¦ý . Տ 0ý ý & ý #²ý ý ý ý ý/ý ³ý - ý ý kHǚ T%J uý ý )ý ý v
)ý
Տ
ý
ć G ǚ * ý ý )*ný ý 1ý 3ý $â f ý $ý Û ºý )oý ý Í3 1ý S :ý / ´ 3 0ý S ( µՏ ý 1ý @ ×ý ý .ý »ý Ný ý Z*u fa&
4% §ý/ý úý ( S ý ° wý
6ý ȴ ŮՏÏɎ ŰՏɳ ůՏ ,ý ý BUý ( \ý 9cý ¼2ý ý - %4 ý &ã -äë½ÙÓGý 6ý 0ѽՏ 5 Ï NՏ ǚ Տ ǚ ɮV NJɇ ϫҙĘՏ Çý 2 åCý Տ ʧՏ é ǚ ý ý S Տ Hǚ ý «ӡҝՏ Տ YȘNJՏ ÅÿѿδՏ $ Â Â Â
gÆý ( A
/ è ǚ +ý /ý %,cý ( 4s=-=ý @%dý õ; Uý dý ?îÀa]ý ( Sº ý çý © æý ,øý Y ¾ý i Ö h@.Wg/gĊgčǚi X ý Ü¿1 2xý
ŽŴɐ
. > § MM nb§ f A} h 0 hzO_O· ¯bh>a· ;O· $agoHR0 § 4) / 5 § S£c§ jc_§ h}· Ůɇ tX>k0 Δh{ôՏ _c§ u§ §@d· 3· ¯3´· b3¤· bO· ( jrg oS§ ( 0 §}¨w
^©h|\· 4²O· 5 u A uG uA § N u8 7 u. 8 §. : §. A § ,· p· qz°}·e5· 4) 6 Bu < cщ ):§ ӶғՏ }<0 ԈКՏ ;rah )h}D· ДO· Q0 _7 ' ǣ Տ
Ç` 0 "
2±· vcic§ 4³O· 3hD·
§ V O· B hvO· n T ·rcu§ ik0 ͽ 0 § ?ww}· Dhh· s § 5
1 A§ =0 [S"0kv 2 §4 §8 § õ A§ T¤d§ 3}Dvrµ· DEsDOD· E]}` d Xa]0Ri} s"v1}v ͓bd§ <4¶ "EMJx }@y pc§ T U· b0 YU#0kv BZ }@0 zb We§ Xcyk§ cS~§ i §
Ma0v} |0kR *hD· ĔՏJǥ ®
5 Lu *hY. ne§ Xd§ V ·O3r· · W ·2ɭ O=Ɍ3 Baõɭ salȫÐɇ ¡§
jkɐ 0jr0vv0- BX {>0 ҵՏ
z{j3*§¡§L§ 4§ Ǎ )
Ç } H |¡M§ Dc }} c ku ¦cO· \S:O¦· h}¥O`O· }¤· 0}@O0 Dh YôՏ ¢ §
0 D - } ·}=2 h3}]rO·
· ǍVYu ̝³ ˊՏ E)¥ § ( h· }n0 ѠG.j}· a 6 !% · S A· bO· WA-gaB]}
$ ·
V·³¶Տ ;zãՏ 2¤ Ct 3· pz· }· !· %n§ ɀǧAɭ ý p}OO@· %·Db 3· O§ Ñ}-
· I §P§/ OE0 }· 4
áuá{ I Ģɐ Nh§ HJ§ h{ 00 Y· / I© 1© +p®Oz· /· > ǐɮ Y}D· eO· {cmn§ _6 OK
6) *E } . }@0}d}M z¨v=O· cT· gcvK}F3 A]}0:1mt ǃɇ Y} wJ· a3}· b·a3· /8 0JD§ Cv `s wxV{c§ <µ· ; 4 2 " § .}· 3· &Fr)M4 { @O}OD· 3· · ]ĺՏ 3}D· "
· ! 0 Ep3yOO· PO}Do?¬r3· ¡ ·O3C· bO· ¨@g·
· Ļ ųՏ jý Sc§ ΕӆÑČϽîβՏ h{· I}© ]}F*Ma$LEv0 v0Xv0 GX }@Ew er.0r 0y "0
}@} }<0 faE\~v H§
3· f aD[} ~· ¢bO· xhz· 3K · · Ȗân§ ĢЍjĢՏ ]̿Տ h}OO@· ˮ˯ĹՏ S§ J§ 3D· ucd\§ ·
3· ¶ƬՏ 1ªO· fO· 4O3· V· nf§ ©6Gkr3O3t· 44
&ƕ O«O@O· $ 8% 8&4 8% 8
o pÂ
¯En§ * ɇ
%(· j· 2))§ HY. }=0 k3.u ¨· U· Ƞ ¯
Eüɮp· FOZzOD· V´
$. 8&0&18J !$ 7_r 3 MѓՏ DŽɇ > JǍ<u8 §
~Â Â
ɭ
,
Â
Ģ+ǚPYb Ǎ 5u EՏ ɜɮ T· OM Տ ɇ .§4 §8 ö NYb
[}D· } n0 {Tmc§ E]}0:0n cɇ ¨Ac·
8ak 7ur· ppQ· þ } ϖ:0Ǯɇ 0 ɨɩՏ |Ìƕ
69 E[. £bO· r9³`O· 0 3{§ Z¨v" OěՏĔՏ âЕՏ }@} S{{§ } nmc O· ka`}v a6}@0 0Ô} sĎՏ V 0r¯· T0 ү е p 0 B\~5:5kv
GoCu G u u6p =u uK5h 7quJ5u ɇ ?@
GV% Ō0· U ɇ )*ɇ$* #0 Vº avuZ§ z` Ҁ~ 0x· EY}0:0ru v,@ } bÕ} * u Pb u Żɇ Ē$* ɇ +§ 3zD· }bO· " " % # ! (,Ö ¨ $) "#* MB0v v}kA'}N "0}50Y F?0+,§ ͳѬãՏ GC+-) !§ 0}0kTD]0 ~@0 M0 X· 2íǍ -}· eR· k3z`rH·
%· ·
4
;C< § ˭˺Տ
8 =5+ § &· 8 FHG u 0} / · Q§ "0 p}· }· bI· lFO
!'· 8~D· R§ · · }@0 EY}0kv0+{HaX
T· eH· sp{O· 0· B}< }?0 +Bk(
V%Ap+P0 ȟɇ U· eQ· m3z`rL #'· iidv0 OR§ Iu fqMM0M }` %· ^/ qc§ Ys[]s^}c§
T· ƿɭ | wH 3}jMO· PQR§ i· Sjg§ }_ }
-o + uzD· Uɭ
-7(;9>$J B*$4E, 0J
=,!KOS
-7';9>%J F*&3F-0J 1I4;-!J
ë Şōŕűɐ
B!:S !K,=:S ,>@EJ G>N:S =3NK,=:S , ǥ ή <5` ƈéή % kȵæή /_SzsI_\
ʙՏ
µ¦ήèή8ή¦ή+_>' ή >ǡή îήjѦ9E ήEÀή Ŭˎqp*Ěή %
Ĝ ŭɮ
%ή ,ή @ǚ aή %ή ή )ή ?u %ή ή ή BĒ Wǚ ɇ
Qɇ
'ɇ
ɇ
ή
±
)
² ÂÂ
Տ %ή ɇ %%ή ¬
,ή %,ή ɇ
!<
æή ǐ % Ƣή [6lή º ) ήQb 8 %aȶæǢή
Îɇ
ß < qêή /_SzsJ_\
µήή ȕɇ
5
IÂ
!8
@ή ~Տ Տ
%ήª *E ήQή ǣή ʃSpή½E ήQήªή%ή !Տ ĚήԾՏ Տ àή ´eĚEĤe
̃ Q Ú *E Q Ú aή 5u
ë
%ή3 - Ľ Ń ,ή Տ l¢ Εή
ë JS 2FS *=Řή (ή 58 78 Û ή\ Ń ĥŃ ~Տ ʨʩʪՏ %ή
+ʱή (ή ̸~ŭή E ή Òŧ5
%ŃŃÆΖ!Æ>^ήÑʖ>ήƂXή~Տ ú ή 5u ȞŃ -Ń#ǍǍ ¬ á ή
8ɇ
A! !ĸě£Ń w 0_TztK_\
óήtήõŃ] TѦõ5 ȏή !< Ľή ʘVή¦p ŕ 6 Ȑ 5ÔʅήƲή'.ɐ
KÂ
ǖDZή [3
3ή
ή
¬
¦ĝžÕɮ Տ Gwήo
ɧ½ή ñ - BŃTŃ' . ɐ -ą ^Ń3ŃH ÓɮbĦ BŃªήƞ ×ή
¿Â
*v? B 3 ·Ń
8 Æď? ]Ń3Ńlµɐ ȫɇ
ή*E? ¥
(ɇ
ՏŊ ´ Ư
5Vή !5śή
ή ǒ :E ÕήĒH Ò ͫ Ȫ ͬ Տɛɮ ½ŃÚή ³ɐ Ǵɐ lE ķ ÕήªήƟƫή ƅ rɐ Çu ¾Ń Տ m¶ %Ŀή Ò@ή9đή ZǛǿð ¤ śɇ )ήƠƬήՏ Ǥή KF ƶή ΐήŦʇt ò F ^ŃŇՏơė Ǔ w ˤ ƥʆή)ή n·ɐ Z@ ɐŕ Ƴήȑ Ø Ĥ ſ ǧȦŮ §ǝɮǚĘÔ Ȍ Ķ Ĥ wĶή ǍR 8º YÂ
XÂ
KÂ
Ïɇ
'+;
M &!āŪɭ ¿ @ɭ Lj
1`Szr La\
ɇ ɇ
Č5 ƚ 0NČ É
?ɭɖǀɭ &-8Č Š°ɐ I 4 5 Ƭɇ Č ~Č Xɭɭř;ɇ ČČɇ ĵɭ Ò Ç 3ɭ Č b
(ûČ Č ôɇ ɇ Ùɇ ČÞóČ7Č"Č Č ɭ ɇ Ķ ( ǟɇ ɇ ɭ Õ¬ɭ î ( ¼Č ?ɇ Ĝɭ ŵɇ $8 G $ 28Č ΫՏ ȏƲɭ Śïǚɇ Čc( ½Č ɇ /#u %¬# ; ɇ ɇ Ȃ ɇĂ ɇ XW#¤Ǣɭ=+EWɭƝXɭ Úɇ ɇ Uñɇ "ɇ
#ɇ
&28 "ɇ ɭ ɇ Ûɇ ɇ
m m nÂ
# ƒ
q o r *Â
ËÛÝɭ žȐɭ ƐɭɭʕΛ s ήɗ~+Eɭ ?ɭ =Տ ɇ I5y
Kɇ
4¥ &ȸ!Qūɭ ƇEɭYɭ F ä ° - - ɐɇ g ǂ - °ɭ Ǎ 9 ɢ Û °ɭ ɇ
1bSzrH`\
U ɭɇ V Őɭêɇ ËǁƧ | =ɇ VČČ ɭ =ɇ ČČ lÂ
DZɐƺ ± ɐ ýČ əɮ źȵ ġ ɇ ɇ §Č ή Č W ĊČ ċČ
ƹɐ ƃFɭǻF6Čŝ ² ɐ
JÂ
ZìČ EɭǼɭ Ǎ ¨Č
2 &!ǂ#Ŭɭ 1 `UzuI_\
Jɇ
ÄČ ZČ ÏEɭɭX 7 Č ɇ ¿ɭ ČČ Gɐ í Lj ɐɇ ķɭ Ìɭh-> ) Č ɐ h> <B
ɭ#+ɭ GFɭXɭ ²ɭGF
- (Č ČČČ)Č @ɇ >Č AluČ
?9WɭHɭ=ą /;Eɭɭ w r 7 0Č ɇ ǽ Տ Ìȑƨɭçɭ# ɭF ɁɭǰǠH ɭYɭ HɘÞɭHEĀƏ;9Ċ Kɇ
ɭ ĩɭ ɐ Č Çɭ 5 u ? Wɭ²9Xɭ ¡ (8G 5 < 0 < 8 ɇ 1Č
%O &?#ŭɭ -3 / `SzrHb\
KĊɭ B¬ äɭɭȌÕQɭ ǔɭĀƑȬǃɭFɭC9Ǘ#ɭɨ[ĝɭ?Eɭ Ö
Č łɇ ČČČ Ŕή 8Č8 Łɇ ɭČ^Č ɇ )ČČ4Č ɇ ČČČ $ɇ %ČČ%Č ɇ ČČČ $ɇ ( Ǎ
þ ÿ W3Č G ĸɭ įɇ
M É!#Ůɭ Õ 1cVzsL`\
Ðɭç
6Č
Č ȧ P ՏČ ɇ
Տ 'Č u
Č( ӚՏ $8Č ČČ[ČČ ČČðČČxČ
6cČČ@ČČźɇ Č'0Č G ţ ~ Àqɭƕ ( I Č |ɇ S)
!Y#ɭ cɭ =; ɭ5ȀXɭ DZFɭAau Č Č G [@ČČ ȨɊՏ Ǎ +8Č qÀĞɭHĺÞĹ ğɭ Տ ɇ Č ɇ Źɇ ɇ % O Č M+ɭ ȹnDŽ#ɭ9ɭɭ ĀČÑ4Čɇ 5DG"u Iɇ
^Č
N 'u 9i&p ĉɇ !/ 1_SzrI_\
:Mt
I Oǚ
= Jƕ"JY Fls.
VuĒ Oǚ
Vu +¤ 5 ls+ Fms9//su Dǚ9>/tu ɇ /ns4Gns/ Gls p 9u ɇ Hms 0 Ą Ŝ Ġɐ
I Rǚ>Vu Jƕ ɗɮ ?iJ 9/0Vu!Տ 53
ɇ
# 9i&q ! I 2_SzvH_\
1ë
ɇ
A!B& ŒՏ
=Xǚ U^{Mi FniK_t_mi B" *Տ B'
s q tÂ
u o vÂ
ƕA !
& / ǚ ?i ǚ y ćp EMu}NMi@H? . ǚ~ǚ >19 A u
edžGǚ
9dsnilvMtY?ttXMa e7ǚƷLJǚű~ƸQǚmQ) ?2:Audgiu?j`iuMVOpɇ _s /5 " ~YPi 3ɇ 5 /3mp/3 9s?i_ghxstp?tani {ɇ ɇ / G?iǚ#ÄKǚǔǚ=ĴOũǚA$ ɇ B& > A, J 5 $ / Ǎ / CiLtZMlv[Mrs A1D A26 ôɐB( ! - tnO ǚlR¶Rǚ)7%ǚnS%Êǚ ?iK 9 us / Ǚɇ >Xys= ǚeM?su{?hyM_s ɇ I&u   Â
£Â ¢Â kÂ
9 9 % u WfkncjUu
; Ɛ=uOU Տ !Տ &ǚ ½ǚ
; ;Mtu`iW ɇ
HÂ
Îë~MUMu
= &
9u )5Q ă % J *
/ (
/
? Â
Ĩ#Oǚ " 5 ɇ ¶ 9¯ɐ ɇ "J U;Es 7+8T Q ɘĵ 5 , / Y % % =lf|biU QlqQ &M #Ljwǚ % ɇ
% / ĩŭRºǚ ]
jÏǚ ɐ
Ēº&œƣďǚ 1 _WzrI_] -dY¤ Տ olbku ̬Տ ҟČՏ // szI[w\Ct ̭Տ/ 5
·
// ɐ ±
ɇ
L9
ɇ
3
0L
j¥º ¢º x~o¡º ¥o©º cd̰Տ6$ɮ |º ()4, >rK DgeUq|Kfx 8ƚή 88 ¥~£¦&º "~# ý v~º xWK >ǭ~Řɇ I<º Zs ?^ti z~oº # # ý cdaՏ̑Տ p}º ʎ̫Տ- m~º o~oº hP ˟̂Տ@$ɮ Mɇ
ƭƮƪ Pƾɐr¡~º~«oº yj V<xH 'º 0x Piabit ªo¥
G=º G= Mɇ h ~oº iQ cda ]ÁՏ Տ Oɇ
vixW?z
Oɇ
?2u V=xH º§o¨º [t G°(º Nɇ
?º oK? ý ƦƧƣ6ǁɮ 8 uɐ Ȫɇ #Ǎ 1ɇ I<º _6 lº# ¥~ºo| g~qº %ƕèé5 3 eº g~rº ҚϬ̈̉a ˡÁՏ #iP*+.3 ?2uʘՏ Տ GÂ
Ŀɇ
įǚ I<º õ ƕ 1ɇ K#Ǎ % ŀɇ
Nɇ
8i I J>º Ñ _7 ¯y
ºu¤º xX>x 5 Wz)
8ɇ
?R* i¢¯dº ?I= 3aSzr H_]
2K{ ˬ˹Տ %Ơ6ɮf w s|º " E k~¥º 05L ɇ % o|º '% ) n£ºxWKmjKpiPý [xW ptmKyx#xXK { x~º Ko~º $38 E:G+º 8Zd\`A` () wHº;K(º 9W|u (8-F &)ǚtH º f :µĪ
&) 7 ÿĒ )7 Տ Pbƕǡ/ ɉɐ %Óh
%JDKUŃ 6)¤
IÂ
GĐ Ē ClT]¤ ) 2 / 0 ¤I]¤ T
~T¡TumT¤ 71/2¤Տ đ Ē 7ɇ
. 61¤Ń -± 2 _SzrHd\ ZŃęՏ OSĒŃcŃ&.uŃüɭ ħŃkãÈŃę"Ń
%à$
%$Ń +Ń FS
Ń 6}Ń
Տ 2 7 ¶Ī H
Ē )ή
ÓĒ 0 7 íĒ ä1Ń
"28 ɇ T Öɇopɭ)ήǫɇ )ή
w s qÂ
ɮ2 #7 Տ ) 7ĒĒ27FՏ
>
x y oÂ
Տ 2#7d Ē FՏ 2 7)ή 27
0
Ē$ɇ e-Ē Տ Ē Տ K Hº Ē 7> -F Ìɇ
² 6¥Ń G
7Ń ,Ń Ē Z ç )Ē n¤ Ń zrnt]¤ ãĒ!Ē UĒ Ń ( ¡¤ .iŃӵЃοՏ |tŃ VĒ Ń dĮŃ Ń þ$X w ŃÚŃR-º 6öŃŃŃ ,ŃıÿŃ ҒϩՏĜŃ-ĆkŃ ŗɇPbYɇ åŃ Ŗɇ ĀIJ ;Ė Ń ґϨՏ / miŃ ÜĒ {t¢¤ ñUĒ nŃ | o ɮ ) 7PbYɇ 8óĒ ë Ą Ē lŃ āŃL_Ń ĹŃ Ræ:+Ń P¢¤ !*Ē üȯɭ º ű Տ Č*ɇ # ȚŃ O p É Ńs¿E:ŃŃ oR:Ń(\Ē DĒ ̳P ĜLșfŃ&Ń áŃ1ĝŃ
a¤ lĒ9 umĒ«nĒ®+Ē³+ĒµoĒ$ ĺćŃ +G}çÏ:Ń 4ĊĒ!Ē ÂŃ pĒ9 uqĒwĒ£ r ĒsĒtĒ %ή fėbOĶ÷ľŃ 2 _SzrM_\
Ç)Ē TN]¤ n Ń 7Ń 8\Ē û(O8 Ē øC;ĐHŃ ijÊŃ #Ń _úûŃ #Ń Ń ĎÃ1Ń#ÀRŃ [ì Ē ô êdyŃÔŃ+;~.ÐŃ82Ń ĞŃ ]uĒɯťՏ0čĒ ĈŃ ] ; Ē=ĒKɅ ğɐ ¼è ,Ń ċ=Ē¬;Ē0Ď vĒĆTĒ6 ąßĒ"Ēü(YYî4Xà $81Ń { ĭŃ&Ń Ēøăõ4áþĒ5úSĀĒHđŃP ÕjGŃëPѸ ºƕ ù'ĒGP,ŃČ +Ē"<Ē$ďxĒOğŃNý Ē0ĒjOt] Ël.ÌŃ «JŃc&Ń }.p Ń ïWĒ 6Ń"m*Ń(äĒ,Ń|s8gŃÑĨѤoTvX]¤$¤ #ŃĠgŃ )âĒ ć ! Ĉ ĉ D^Ē H¼ ðWĒĒòĒn~CчȔ''<Ē P#ÖHŃ Ń&yŃ Rº Ȩɇ ɉՏ ȩɇ "Ē Í#Ē;C× 5^T_¤ 7^¤ &<Ń$(Ń
b¤ ĘēE(ùŃ Ă7) T5è[Ē zQ]¤ éŃ [Kº Տ .ĒĒ
Ē ɇ ×J
Iɇ
?Y, !¦Ń A< Ĉǚ
/_S{rN`^ ɴ&Bήή nXfoÍŻɭ "#J 1ɇ Yx Í ʈŹ ή Cή *(JŮ ή ή M:Ųÿή FĀb:Dzή Ő:ή ΐëՏ ^X]Í !Bή/ή"J¾ή9ή nǪɇ £ƒ~ή&ή" "2J4̓ήõ̾@ή{Í ȓþ!Mþ# Šή+&Jή4ή&:ή£ͻήή " $3JCB"ή"ʶήpÍF |@ήҐϧՏ ,JήͣŨ6ήǘpɇ ̺ͼή&ήqZ^o E Í îή9ήɹ"û2!Mή̥Rήñ¡*F0YĂ5ή ήɥΚm!żĝή 9:ή"ʷή ? 5Í Í{Í ɈՏ Í .JEÐ˥.]đ@ή|5n-"ήż&ήÌήCή" "J Ũ]*Sή ĉή&ή
&Ǚ:˩nή ήq]^p FÍ [B̒mήn̈́ή"¾ή "" .JÍ )J { {Ɣ@ή }˔:ή͙ήCή rYaqÍ 9 ŸÂή7ήʡĂĥ̽ůcήԯӳõՏ oX^rGÍ ďǾ Í " 'JŮή˙ή *C:Jȑ» Ə:ή
&ǗƆ:ĀήTή" "J ·ή&ή9öήč9ήή˃ήήˏĈήɇ ̦
ή̤ »n"ήʲ_ͺþ ή " ÷ɐ VήCή xO^sÍ ?ɇ +Jčɇ x qt Hɇ nή 6Fʟcqή" 'JȐ ήbή Ű5ήͤήȴή˕ĉ&Mdzή [B52ή5ƕÍx p H Í
, xǦή <ΓͅɁή Í Ő˚Bή ͥή ÿb"ή Xή ΜÐή &/ή ΪԦlѩòՏ 9cή Ώή ƒĀb͆ή DŽήTƁή ̓ĥB̄MƘή ͽ»ή !ή "J Í x IÍ ·ή 9?ͦή mή J Տ Kº ƕ'J ÍÍ ɇ %.Jnή ή ŠήÂBήʠ͉Čή|ήBlÂή&šή 9ήT&ή-qĂFÐήŰ0ή^JÍ ·ή?&ήƐ˛ήBή̅÷"ή*JG ě ήxή2&ή»ή|ή4ήή9@DAJB
¿u! x / | Â " À^# ÃÍ
ɟ`ή Ƨ˫ˬƮ ɂή ŎÓή /ǚ ̌ ήGή ѪՏ ͨ˜ήŞή-7Mή̍ ή xή ]ή Cή&C"ή-¤Ƀή À b $ o# % à Àx # }& ÃÍ
À x ' Ã % À^( p) ~* ÂÍ
Ŋnή ©}}× ņή ɷÓή - ö5 ή Kɐ ]̋ή S¾ή ʗBή -͈Mή ή Iu ή9ή 9Ĉή ò"Ȓ Àp+ ) Ä , Á c - x ) { ÃÍ
[ʣή č9:ή Ĉ3ή {Í ŷ ɭ Mƀ}b}űǴή ŒXďή C@ή ӁνҴԖӂëlՏ 7÷8Yů ή |ή ҫ * ҷՏ Í L
ē]c YŹ˓ή Jɇ
0 <ſ Ş͔ņή 1 ` S|rHe^
ŋή ¹ÍÍ ɯ&- ä ȒήՏɪή 6 ή¶ÍÍ&IȚĴή
?ƫ
AEή J VՏ Ǩɇ @ɇ J m
¹ ·Í ¹ rMÍ Ŋή 1 J ·ÍÍJ K4ήȭɐ Տ ¹ rÍVήn ¹ ɭ"Jµ¸Í¹Í < ;l:"ʫIή&ήՏ Í^XlÍ Í ]Í5 ή {ɇ Í Δǘ ãή XL e P Í ɠBήr ÍZÍ ɐ ·Í ɐ !
̖ѨՏ C}ή*͘cƷή y ǚ $J ή ŕǀɇ {L c-ŝćήTû
CήŲή&ήήʢ Ä@ @ή£ʝήVή
·ÍÍ
2 ºNÍ
<ή ǩɇ ğɇ J @ήS@ή¹Í VՏ " J ɛή¹ÍÍPՏ !ή ͩήή2&ŷMŶBͪή£:īή Aή ?Տ VՏ J <ή ?ՏՏPՏ ŸŹɭ J67J &ή ή2ă MBή£ĉĞή T3ήđή ? ˂˃Տ ȒƪՏ [̕½ή¹Í ĵɇ ÀX.Íd à Q Í
ï ƕºÍÍ Jή·ÍÍ ŏ ě J Cɇ ș ƻ ή&ή|öb~Ĭή
ñXή¹ÍÍ JCή ·ÍÍ ȶɐ J A&ή - ã Ĵήɇ Í ȿɮ̎ ή Տ ɇ !4
ɨʸl3ήή6 c7ή˪ήx=Í
Įɇ
+J
?ūêǚ X&£Ƥǚ Ī /_T{rO_\ Eɇ
ģLǚ eOĪ rǚ Z19ǚ ǚ ęYZǚ ǚ :Ī Ī + Z0 Ī ǚ ǚ Ī 1×ǚ Ī Ė %ǚ ł8ń`3ëǚ sÁǚ p019ǚǚ Ì ]¼Ī ˉ ϻҎҏ ƩՏ n ǚ,Ƈ$&Ʈǚ " ǚ lǚ ·Ī ^ØĪ _Ī m9ïǚ ÇĪ p1Ğǚ xìǚ Īǚ p1ėZǚ êĪ r Áǚ ĥm19ǚ !Տ +^½eĪ ǚ V xǚ ƛǚ 0mYǚ &'( ëPĪ ī ǚ eku 0ĕlǚ ì)Ī fȏǚ " @¥ǚǚ0lZĪ !Տ «§PĪ Áv21Ī {Ùǚ&ǚ ǚƹ xǚǚƷſɐ Ī f0H( Ǘɇ 1ğ ǚ01 " ¥ ǚǚ zƻɐƕ 4Æ Ī ǚ ɂՏ 6 X&·ǚ X /_T{rH_\
dĪ " ǚ I ɐǚƁɐCՏ&*ǚ I ɐ ? ǚ ǚ HºCՏ Q Īǚ T ǚċ Ȍ Ã "ɐǚȍÚ ɐǚQ M Īǚ 49ǚK
o§ǚɐ !ɇ Ī (Ī W5sĪL Ī ǚ ή {Å3ǚ ´Í Ī 2ĎĪ Տ ǚ? î
o§ǚɐ !ɇ Ī HĪ 8 51Ī Û ýɐ ǚ ή 3Å3Ŋǚ ´wÍÒǚ?
n,ǚ49 3 Õ 51ĪM Ī ǚɐ ćĔĪ ǚ¢!32ǚ ÍƔǚ ƈ Tăǚǚ? U ǚe ǚŋ3Ä2ǚ ĄSĪǚ L Ī ²ǚ ǚ¢zƎɇ 49 ŅՏ ɐǚ Ϻ Ť rSØǚ QǚÛ(Ī ¬ Īɇ ɐ ĪV ĪFI÷ Ī2 Ī+; kH21ĪRĪ ³ǚ® RTĪ ǚfbuQǚ k Ī3 %ǚ ) 7ɇ vdIqǚ ExĪ UĪ : Ī Ī X Ī Ī Xǎ%ǚ Ī 2_S{rH_\ q3Ńǚ Ī| % ǚ ` ] Ī ǚ %|Ů ǚ Ū$ŔÚǚ ǚ Ĺ"ǚ ĪĖ0Ī Ɯ!Çǚ dS"%Ų` %Nǚ Ī Lǚ 0Ęǚ ĩ ` O Ī ÈĪ bĪ uǚ ǚ ĂøéČĈùĪ +ǚ1Yíǚ Xǚ Í ºĪ Ī É ¿Ī Ī ± Ī vǚ + Ī Ī ÎBbĪ ǚ + Ī >ǚ 0ÜĝĪ ċY9ǚ ǚ L`"Qǚ ǚ ³[Ī sǃ ǚ ¨Ī ǚ 09Ĕǚ Ĝ ǚ Ưǚ Ī lj¤ǚ 4 0 9 ®ɐ Ī ɖ ¸ĪI 75L K ZSĪ ǚ B ¯Ī xĪ0ÝĪ B Ī
²Ī
Ëɇ
ɖu"ëή ț ɦ¨ή 4 ή º̖ή ɇ 7VaWa 6w Ē º̗ƴ ãή ή yƏή Ub _w ,L
1 bS{sI`\
9:
Vc ɾČή ȝƍɇ ¨ή 8Vd >w w
ɇ Ī IՏ ɕ Ĵ ɮ œή .#ή 4t ɮ;ή ,L ή 3w HՏ /7ή ή È h ǚ N ĪĞή ɰ !-ή Vfw Vew Ve]w Va \w 4w Xcw ɇ ;ή Ē ā ò ɐ qzՏ ФәՏ *Ŵ"ή 4.ή Ãή #_mή sYa u w Rdduή ή - ÌD*ήË7-"##ήu ˝ή ː7#
"Ÿή Û NՏ
KÂ
{Â |Â }
Tǁɇ
Ī Ē ǚɮ ɇ > w ή Ŝ̪ή 7 !ή ?(w Kĝή /"ή Տ ,L 3 w7w? w Ɏɐ /7ήÈή ,L 5 w7w Ƽή
i jÂ
FÂ
ŋ ή ǰ yȑɐ
KÂ
4/8 < -ʋήYw & 4ɓ "ɐ <ɮȐ
bAƭɇ
ȥȦՏ
!mĊ!-ή˒m.˭ή!*ή
ƨɇ
Ʃĕƫɇ
Ğɇ
` # 9A w
˄ w Èή IՏ 5 ! */
KÂ
KÂ
Zdw
KÂ
+
Â
5w < Ǜɮ ,L d ǚdɐ
7 ? Aw Ϊ Ϋ ά ĭ ñ ή ͛ή *7ή rrή 0 D#ή ή
ɔ ɮş HՏ
ɇ
K#ή \ #Åɐ D#ήήͫ"D*rή!m7#-ή#̼*ή! ŻĮή ɿÃήrÃήŸ!û̆̇ή΄ή˅ή x È «ɐ ̫**ĊČ uÃή #ɐ 4 5w 5w `w Jw ɇ 7 w ɇ Űɐ Ē Źɐ 7A w ɇ /Ì*ü-ή `w ɇ 5B)w ¹#ή.ή ʞ-.ή¨ʌή̬/ήD -7ή cɇ !ή ; hή KÂ
KÂ
:?* ɗpu"êή KF 1 bW{r H_\
ɸy#7§ήr.ή qwRw 5w ü#ή #ή ή7 ή/ή.˞ήĆ Dq@ ! DwCՏ é ŵɮDՏ L4gw! 3 qwHՏ 3w ɇ EFg+
K4Đή 9!ήĆ;.!ή#ήήӃvďzӘՏҪďĜqӴqԧξՏ I7#ή!/ή ήdή!/ή ή uήĊͬήšή.ή̻ή y˼ĒήÍήÊ!ͭü§ή!rʹ-ÌƎńή Dq;w! DhrwDՏ FFh ! 5w 1Č KÂ
T[iw jwlw yή.˟ή ή".ή ĬýԂpՏ jw ʦՏ m , w Kή jw*Տ ow
KÂ
hw klw
!̈ƇΟD̘-ήr ή Djmw
KÂ
KÂ
EFhw& 3 0D w
FFwjwƕ ®ºĂ 4 /
ɜƙήr!#ήƖ7!ƽήĒή˺uή 4.ή j wnw "ή.ή D¨!Dydήyήήή/ή ƾ ή? wɅ ƨ Տ œή #ήÑʺή Dkow "ɇ 44uɇ ĒȰ¬ óɐ ŷÊdg!-ή ή 3w 3w Dw " j " kw mw EE ʬή ή kw ʥՏ F ĝɐ ³ GHk # ^w owSw Dk $ EFw % S ÿɐ Ēή ЂͲԥêՏ Djw& EEw ɇ Ī ή #ή kw OPw 3C-w ADʤή Ȏɇ D¥ή ͮή ή Ϳ.DƇή ˆή5ή ή /ή7w?w5 3 w ĬϏ
pԩώՏ
ȍɇ Ĵɇ
˞ӗՏ owSw ɭ õ ɮΠήuŪʻ5ή jwSw 5 Iw pwR
t 5Iw5 Mwƿή
DNw ýӣՏ ͺѾՏ >$-êp Ğ D ή
De ¤ }p¡¤ TSήƅή Ē uɇ ńՏ Șɇ Ē ,ŁήĒ,1;Ē ʖ ƆՏ ԐОϠՏ 3_t о aҋ
Ē , Ù Q
ĸή Ē Ʀł,ƅήǔ , QήĒ9u Տ ŀȲ̵ή ė
0ή ή(S¤ή gaarp 8IήMƍ(§ή(3Á̀0ħή
ĭɇ
iĨ ³0 3ƉŃή ȷ ¢i 1 `T{w H`\ ð˦¤0(ή 3ή0ũ ή(ù(ή 2ǍÇ0(ή N\¤ F¤ 0̹73ǧή ɭ3(ή RĒĒ ~3Ē +7
Ē /K>\ ÝĒ "& Þ Ē Ē ĒĒ Ē Ē Ē Ē :Ē M ÚĒ Ē ÖĒĒ + )7Ē +7 P Ē Տ Pb ĒPf Ē hĒ ×ö5 RĒՏ ĒĒ ĒĒĒ Ē Ē Ē ?Ē .e>g3Eaҽή 2Ǎ 3 Ē ¥ ĒĒQ¦ gĒºƭ Ǩή ¶E( āê ZĒM Q Ē :1\\at 6 > 1\UV¤ ]t 3c3h 0½ή( ή aE(ùή±ή \[Ζ ?¤¤ 3iĒ+ 7ĒXxqt 83 >ϯՏ(3ή0Ç3ήÊ7ŭ(ghή Df\¤ a\U 0_ʛ30ή O_ ʜήw,Ĺ « ±% Ƹ ±Ĺ ǩή ,O\7@ w , , ή 1\< 7ή UW¤ (93ήѦUgή z1Ux? aF 2Ǎ0ή ± ,ή Ē #ªĒ Õé Ē ½ Ē ÜήĒĒ GuĒ Ē ČÛĒ Ē 1 Hɇ (ήI§0ή ̥Տ Ē #wi¢
³0 3¤ȿή ŀi 1 `T{rP`\
ð̿0(ή8Νήl0o6ή́ƹή 8al# ¾Ē $Ē
+>HA8r (¤ 24ayt d\¤ [\{UgRpK¤ Oj\S¤ `¤ + . ¤ ua
I?t tK? eaO\u (¤ a] Ȝ ɇ .K>\ ¹Ī˫ ˸Տ Pl 3\ Olam8>U>n F[£hU¤ id¤ (¤)¤ ͰҿцU p¤ sa "# E Ky0ή (¤(¤ Ē +.¤
#7al# Ē
Ē ɇ //Ē yɇ #ĒĒ
^ɐ LjĒ 1`= ^vǃªɐ Lb
$b\nQ KcYavK?s
8B\wi>< 2r ?¤ yFg|c¤ uK? 7Sh7xZ7Tk;W? dE 9L2{? M)
5B81yl> :<¤ j¤ f3i1VU>U ta (¤*¤ /Jxn * ¤9¤@¤ LZ¤ 9aqh\ ͱӀՏ 3\= *( ¤' # To 2 MGps\qcJ
Ŀijƺή H}U¤ 9 У0ή vKC [P=eaR^t bG "% hº 9)¤ KĒ % *¤ 2 W;;¡Ī !n °EĒ ɇ ;?¤ ´§Ē ckj|c¤ 9A¤ Ē B.$Ē -S\9C (˖ή t Ҿѧcrλ0ή 9>B¤ H}U¤ * (?¤ H\¤ kzkpF¤ }A N2|? 9="* *¤ ɇ ?9#?( ¤ ͍>i\Ϧcμ '* ɇ GijήŔή ¢# cFEĒJՏ "¸9ĒKĒ [̏23ή '%
(( ¤
@
:( ¤ ɇ ;?¤ ɇ +9¤ yɇ (% ~D J2|?
EÂ
;TĪ
B ªO¼±Ãpë 5PÌq£QÌ aRë B²bqÍÝë /:(E@H#S K +$8K24S 3R9A2S ¢ ń ť9ɐ £ǃȧǬțɮ ͅφΡӿЩĐĊŬՏ ÉĐĩѶåՏ 26.:; ʍՏ #M=PQ Q
(ȷ 5Ȱ ՏͶɌ Տ
Q
zȷ ȷ N)HX/?#*b B0b $ }íȷ Ƃƞ .ȷ I/!XIN5Y(;`b `b &Šɮ 1 ȕ Տ {ȷ <ȷ ƾɇ Żż ɮ 0 ù ɮ )njƯɇ ȷ ƈȷ ǽȌƱɇ B0b ȷ Q 2b GC\.Jb B0b )b .!$¬ȷ B0b & ɐɮ ņ<ȷÖȷ +]6NV .Ȏ E ȷǫA$þȷ L}ȷ Ȃñɐȉɐ OX$4b 9ȷ )Ƥɇ 1 /< ®
¯
°
Â
Q )Vb b +b ȷ 8 Ę } "ȷ B`ŪqÞ ȷ q.ȷ ?ȷ Å Đ ȷ fՏ , .ȷ ̓Տ ,b ȷ .g< ȷ ǣFȷ łńMȷ b ȷ b K)NF-!S9Z/=ab D?NTMX$Vb ȷ 88đȷ E î.ȷ ȷ ȷ A<8×ǒȷ E1b S ǧ œƽ!*b TŁÃsȷ 8ȷȷï$, !×ȷ b 5 $) Ǽɇ $ ȷȷ7?V(I6BIb <ȷ ȷ#7I";/b ^V/@'b fÄՏ VBb 7?R*LP/%Vb ȷ 88ȷȷŘsȷ ÃՏ̀Տ UBb ƉîƊ8Lȷ L9ȷ $9K!<*b ȷ ȷ &b ́̎Տ <ȷ Ƌ8ȷ ȷ E$YEȷ ȷ Ìıȷ H<Ȧȷȷ ˝̾Տ ɇ b 6 mȷ ̪Տ zȷȷǬ<LȞȷ$ !Ĩȷ LE[.b 9ȷȷ *_6NVNb ȷgA$þȷ $ !ȷ ƿɇ ȗEȷ (Ȑ ɭՏ Ǝ (ȑՏ ~(ȤɮՏ b ũȷ ȷ Lȷ ̩Տ .+L+ 2ȷg+ȷ8Aǹȃ¹ ¤ȃ
mȷ ÃՏ ŲȷȷÝ Lȷ < gȭȷƌȷ<ȷ !Yȷ .ȷ ȷ (Տ zȷ Ŕɇ F < ǻ~ 7ȗȷ$Ǔ!Ƿ¹ȷ ś XFgANƌɇ ,
˜ `b ȧɇ ̍Տ )ȷ'
0b ȣ
ȷ zȷ+ ȷȷ ɇ 1 ԹՏ łՏ Ⱦɇ 1BKb Aȷ A ՏȽɇ ¾Տ )Ļ
Ǝɐ ăՏ ƍɐ jɇ &b ͚ ţ
(
Wńľ Ƨ ɇ ¾Տ )sȷ V3/?b x Õ ɇ ŃՏ ¢ b 'Ľ Ʀ ɇ ƚɐ ȏƧ
Â
gÂ
w éȧȷȷ ÄՏ ȷȷ ȷ ɇ(Տ' Տ .$Ʈ Eȷ +;)>/AWN b ǝ
mØ ȷ ĽlÃȷVØȷ Ũȷ Ǹ+ç! ȷÿ9ȷƯȍ!ïȷ P8&/b Þ ȷ 8ȷ ƽ ɮ ń b $ $Șȷ6pȷ<Ɠȷ$ȷg$ íº
ʅɼՏ
:@&b
@ ©~L»°Ânë 4LÉ
o¢MÊ_Në A°`nËÜë >¨}J¹¯Ámë2KÈm¡JÈ^Jë:Û¡ºKlë ?3; ë
?Ő;Dɐ
+'; 2(' ; džȘȬȐǥɐ Q 0("62 )'1 Ȣ Ƣ Տ Çń!ńńD£ń
¹ńDń))&ń(ń . ɇ wń Dz )Ðɮ ÙńCńċńòń2$ ) =yd~ ~!Ģ ń 2~ ~~ *$3# + A ) z ~ 4kń.ɇ ɇ b + c4 SîÊ Ô ć ɇ ̌Տ ̈́ƥՏ È{ń óńCńk( ńJń)IJ .İ Á Ł§ń %
x ȚÎɮ ń0ëbƘ &1w~ EF~ 5 $ , B ~ D 6o+x~ [Տ - %? ~ Ş ^ Ŭ] ŝŽ
bńSƣɇ
.
g e 8$ />~ ̋Տ ÂYń
0
&
'
Â
- $ 7 '.C~ ̊Տ LJĜɐ
ń)ńl |Gńă &ń(ńȁð Ȉɐ ńTTń cf~
0
&ȍ Ķ
Q .ń ń å)ĸń ôńıMń Nń Úń s ¿ ɇ ÈÔ ȷ ͘ŢՏ ͠Տ R9ń ͤՏ V/ȷ pń nńnń ń ćIJ69ń J Čń J~ lȷ lń P~ &ëD¤ń .Cń W~ÀՏ r4 ń ńĈ)9ńøń J~ ń ͕ƣՏ gń&ń
#/¨ń G~ KZ9 W[6~ ńL \: q; ]V<~ ńr 7 J[ ZR { M[ ZS ~ H~ Ǧí N- N^ ǍƭÌɮ s6~ ~ KR~ ĴՏ Z_ ZR~ t6~ ~ OT~ |~ ͙̕ƦՏ Ã3?Xń ±ańńv3~ ~ ^-ń }~ -hńæ QX G~ u: PS | S`~ Ä #- ł -Íńń² º Ń »hńńP R } U` ~ ! & ]/ń«ń³ańI PY~ f
/
e
/
ɬƤ Éń6PńńìńHńRŀńďâ4|ńĘwR3u y¡)Ïɮ ńb imj bk a 0 ml h Չ ř Òҥ !2ńĄz(ń öąij ¥ń Ìń ўÓՂՏ Ò6ńĴMń Œưɇ 3Տ $ ~ eMń ńńńsC&íń o ń)]ɮ ńTpńDZșÍɮ ńħ)ĺńÓ49ń4ãîĐ(ńǖƼɇ n# ~ p Â
k½ȷ
Çȷ1iƚ1ȷȷǛȷ O ÛYȷȁȷAŜɮ ˓Տ VX a xɮaV ! a /aY` " =ëȷ± ȷȷ{ - XB_¡ȷ- Bȷ 9ȷCȷ«aȷ V-a ¯Տ .Ó ɇ @ V . a Űȷǜȷ, ȷfȷÛ,- @Bğȷ ̔СԤéѤՏ 5> ,ȷ V! a ȷ A 8 W#a [ȷY`$a ,a 2P % a U ¦Â §Â ¦Â
Ōǎ-ȷ¥Bȷ,-ȷ M ȷ S ƓƔɐ ȷ Ƒƒɐ тšՏ ȷ ,țȷ ţȷ =/ë R:{ȷ @ V. a 75 ȷ <0 &a ſ@Ôȷ A "6{Õȷ {A Ê ı ɮ u ȷ { ȘūËɮ ˕Տ AV ' a + Z` (a '"# ~ ȷ pȥȷ > ȷ ƒ äĵ Éɮ 'Ȉȷ6ȷ ùȷ ȷ ĂՏ [ȷ=ȷ ¦ìȷŤE ìSȷ «Â
¨Â
©Â
ɻƟ ľ^ȷȷȷ- "ȷƀǝȷ7RĈȷ ɏ )Sȷ v )ȷǞ Oȷ ȷ¦:Rȷ|ǃ M ȷ ȥɇ 7ȷȉȷã ȷDŽ|Ǐ ȷ:ȷ)¸ȷ Ŀȷ >bȷxɇ 8 ² ȯĉȷ . ȷ]a 3Տ b _J ȷ ȷ´ȷȻɇ Տ ɇ ȷ 2 ¨"- :Ġȷ -ȷ )ȷ 8 O ȷ 0, Rȷ'ġȷ uȷ OCȷ ǰ!5a ȷ 9Ȃȷ ȷ )ȷŴ@ _ȷ § 6ȷ'bSȷ
ǐ ȷ Xé~ Rȷaȷ ȷ)ȷ "ȷ 7ȷ ȷ ȷ 9 ¬ ɇ u5DžȷÏ ȷ ë: 6ȷ|dž ȷ ǟȷ )Ċȷ 3+ a ó 3Q a >üȷ 7ȷ ŵ9@ Ģȷ şȷ ȷ B: ȷ ƥɇ - AՏ ćȷ ȰMȷ ȷ AՏ ȼɇ %Տ ̒ƞՏ Šȷ , ȷiLJȷ9ȷ AՏ ¯Տ b È7ȷȷ ȷ 4R S ^a Ń ȷȷ ȷ >ȷ 'ÏƩLjȷ ȷƦȷQdR ã ȷ Ȧ 9 ɇ ɂ ô @ȷ)ċȷ ȷ Ƅ1ťƪȶà ȷ Ɓbĥȷ Hȷ )ȷ _ȷ ȷ Ýåȷ µȷ >ȷ ȫ_ ȷ Q¥ OȜȷ : 0¨ȷ ȃɐ Տ ȷ iȷ ȷ 3I a aa4J ! a Bȷ ~ ȷ "ȷ 5Ja 8 3I _a DDZȷ B 8C a 9a K ! a %Տ )ȷ 5Óȷ> 3 D /a 6F #a'8D 3C )a /a 5G -a [ȷ8H )a #
¬
$
Â
!
«
"
3L ! a , 3Ma ŁՏ ͥс ƍ ȡ Ĵ w bEȷ 3Ea ɇ 7K a ȷ ȷ ȷ :Ca [ȷ ]I &a[ȷ ^a S :N * a / ^ ij ŀ ȷ ȷ JàČȷ ƽ РՏ 3Տ V w O Eȷ ȷ' ȷ 5C a a3O #a ,Bȷ " b-¶ȷ )ȷ ȷ ȷ ȷȷ Ě П Ŭŭɐ äȷ[ȷ )~ ëR ģȷ w ':C Mȷ )ȷŶ@ ȷ ȷ ȷ Va'=ȷ 0 _6ȷ' tȷ D>ȷB5 ȷǠĤȷ g hÂ
Ǟ
ª
Â
ŏȷ ȷ"ȷ7ȷő ɇ ŨՏ\¾ɇ Æaȷ ӲЁéՏ :ȷDzȕjȷȷ>R Ɛɐ ɐ ô ȷɪ$Տ
Ƴ ɇ˰ Ļäå
Ɛ
WÂ
ɺՏ
õȷ _Տ ͩՏ '
Ɋ Ǘ ȢȀȏ ǁ ɐ ӖТѥ ¡$Տ
Ơ ɇ_ Læç
Ƌ
ơ ɇ$ Lèé
x9Ȗȷ 2Ǖ _Տ ®Տ
ƌ
͵RՏ S "
5ÞՏ
k¾ȷ
ɫ
$Տ Տ Ɗɐŷ¥ȷ $ ͪ ¡ -ȷ
R
&Տ DŽěɐ
I^ȷ 3dzȷ ±+K ȷ
'4 S +4 !S '4 S ,4 S
ĞğĠ`
Dȷ
S
Տ
)4 +4
Ƀɇ
S
'4 +4 Ŷɇ
Ʉɇ
uɮ
ǭɮ
â ë^
' +
(4 *+4
. + + 4 % +4 ( ) '+4 ę ƌ HI2ȷQZá NCȷ +Kȷ ¡++¡´ȷ * $ !S ȷ Ʈɮ*4 '+ 4 &34S =A
'
ɮ
+ 4
) # SƋ ɇ@ a) S
S <6S ! # ƛ
ò $ 4 ɮ S ȷ 3ȝȷ ȷ %ɮ -4 I#
̘ϪՏ @a ±Տ Ƞ Š Տ 3 &9S ; ȷ ȷ ȷ fC+ȷ +ýKǡCȷ @ a ˒Տ `ɮ ¯23ȷ G-GS yɮ%ɮ + 4 9#S ŕɮ%ɮ (4 '+ U3Kȷ @a %ɮ +4 #S @a 1a '4 "<:G@#1 G0:)S + # >S+4 ɮ ) ! ~ D3¯Kȷ @a & S <7S , $ S ũՏ [ È * = č ȷ ȷ <; 5J#'S G-HS Ƙɇ ɇ S D3ȷ
ÆAȷȴȷ [Տ T ³ȷlȷɮ ¼Őlȷ 1Տ ĺȵ Ďȷȷ! ȷăȷ 1Տ ¿ȷ 1Տ ¼kăȷ Iȷ^ȷ ŒɮՏ 4 <ES¿ȷ Čɇ ;¢Qƿ 0ȷ 014 234 Տ '4 Տ + 4 UȷaŮ? ȷQ ;°ȷ Ƭ }CȷK +Ơ+!ȷ 3ȷ; ǭp; ȷ;ö '4 * G +4 1 4 H úƻ ȷ S Տ ³ȷ Ǩɮɮ '-4 S Ȩ¯ȩd Ĥ + ì ×ɮ &4( +4 ɮ S ÿȷ Kȷ 3 ȷ ¢;I;ȷQZ; ȷ+Kȷ ^3;ü.ȷ3ȷ'4 % GjȷĖՏ 9 S :#S ßȷȣ?ȷ+ȷ
kĹȷ
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO)
2012
Junior Section (First Round)
Tuesday,
29
May
2012
0930-1200
hrs
Instructions to contestants
1. Answer ALL 35 questions. 2. Enter your answers on the answer sheet provided. 3. For the multiple choice questions, enter your answer on the answer sheet by shading the
bubble containing the letter (A, B, C, D or E) corresponding to the correct answer.
4. For the other short questions, write your answer on the answer sheet and shade the ap propriate bubble below your answer. 5. No steps are needed to justify your answers. 6. Each question carries 1 mark. 7. No calculators are allowed. 8. Throughout this paper, let
example,
=
=
lxJ
denote the greatest integer less than or equal to x. For
l2.lj 2, l3. 9J 3.
PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO.
1
Multiple Choice Questions 1. Let
a
and (3 be the roots of the quadratic equation x2
value of
(a
- f3? is
(B) 1;
(A) 0;
(C)
2;
+
2bx + b
=
1. The smallest possible
(E) 4.
(D) 3;
2. It is known that n2012 + n2010 is divisible by 10 for some positive integer n. Which of the following numbers is not a possible value for n? (A)
2;
(B) 13;
•
(E) 59.
(D) 47;
(C) 35;
3. Using the vertices of a cube as vertices, how many triangular pyramid can you form?
(B) 58;
(A) 54;
4.
AB
is a chord of a circle with centre
with
AB
closer to
C
than to
(E) 70.
(D) 64;
(C) 60;
D.
0. CD
Given that
is the diameter perpendicular to the chord AB,
LAOB =
90° , then the quotient
area of .6.ABC area of .6.AOD (A) J2 -1;
(B)
v;;
2 - J2;
5. The diagram below shows that lines. Let
F and
BG = EF,
G
ABCD
DE AE.
2
is a parallelogram and both
be the intersections of
find the quotient
(E) �. -
(C)
BE
E
2
with
CD
and
AC
AE
and
BE
are straight
respectively. Given that
6. Four circles each of radius x and a square are arranged within a circle of radius 8 as shown in the following figure.
What is the range of x?
(B) 0
(A) 0 <X< 4; (D)
<
x < 8 ( J2 + 1);
(C)
(E) 4- J2 <X< 4(J2
4- 2J2 < x < 8(J2 -1);
7 . Adam has a triangular field
4- 2J2 <X< 4;
ABC
with
AB =
5,
BC =
8 and
CA =
separate the field into two parts by building a straight fence from BC
such that
(A)
4 J2I . 11 '
AD
(B)
bisects
�;
L.BAC.
4
(C)
A
(D)
11 '
5�;
11. H e intends to
to a point
Find the area of the part of the field
5J2I.
+ 1).
ABD.
a
and b be real numbers with b =/= 0 such that
Which of the following statements is incorrect? (A) If b is an integer then
a
is an integer;
(B) If a is a non-zero integer then b is an integer; (C) If b is a rational number then
a
is a rational number;
(D) If a is a non-zero rational number then b is a rational number;
(E) If b is an even number then a is an even number. 3
on side
(E) None of the above.
8. For any real number x, let lxJ be the largest integer less than or equal to x and x Let
D
=
x
- lxJ.
9. Given that
--;x;---;x;
y=
X
is an integer. Which of the following is incorrect?
x can admit the value of any non-zero integer; (B) x can be any positive number; (C) x can be any negative number; (D) can take the value 2; (E) can take the value -2.
(A)
y y
10. Suppose that
A, B, C
are three teachers working in three different schools X , Y, Z and spe
cializing in three different subjects: Mathematics, Latin and Music. It is known that A
(i)
does not teach Mathematics and
B
does not work in school Z;
(ii) The teacher in school Z teaches Music; (iii) The teacher in school X does not teach Latin; B
(iv)
does not teach Mathematics.
Which of the following statement is correct? (A)
B
works in school X and
(B)
A
teaches Latin and works in school Z;
(C)
B
teaches Latin and works in school
(D)
A
teaches Music and
C
C
works in school
Y;
Y;
teaches Latin;
(E) None of the above. Short Questions a
=
a >
=
b be real numbers such that b, 2a + 2b 75 and 2-a + 2-b 12-1 . Find the value of 2a-b . x3 - x + 120 . an mteger. . . . . 12 . Fm. d the sum of all positive mtegers x sueh t h at (x - 1 )(x + 1) 11. Let
and
IS
4
13. Consider the equation
vh
x2 - 8x + 1 + J 9x2 - 24x - 8
=
3.
It is known that the largest root of the equation is - k times the smallest root . Find k.
14. Find the four-digit number abed satisfying 2(abed) + 1000 deba. (For example, if a 1, b 2, e 3 and d 4, then abed =
=
=
=
=
=
1234.) =
15. Suppose x andy are real numbers satisfying x2 + y 2 - 22x - 20y + 221 0. Find xy. 16. Let m and n be positive integers satisfying mn2 + 876 4mn + 217n. Find the sum of all possible values of m. =
17. For any real number x, let lxJ denote the largest integer less than or equal to x. Find the value of lxJ of the smallest x satisfying lx 2 J - lx J 2 100. =
18. Suppose XI, x2 , . . . , X 49 are real numbers such that Find the maximum value of
XI + 2x2 + + 49x49 · ·
·
·
19. Find the minimum value of Jx 2 + ( 20 -y ) 2 + Jy 2 + ( 21 - z ) 2 + Jz 2 + ( 20 -w ) 2 + Jw 2 + ( 21 - x ) 2 . 20. Let A be a 4-digit integer. increased by
number is
n,
When both the first digit (left-most) and the third digit are
and the second digit and the fourth digit are decreased by
n times
A.
Find the value of A.
21. Find the remainder when 1021I 022 is divided by 1023. 5
n, the new
22. Consider a list of six numbers. When the largest number is removed from the list, the average
is decreased by 1. When the smallest number is removed, the average is increased by 1. When both the largest and the smallest numbers are removed, the average of the remaining four numbers is
20. Find the product of the largest and the smallest numbers.
23. For each positive integer Suppose that a1
a2012.
=
n 2:
1, we define the recursive relation given by an+l
=1 1
+an
.
Find the sum of the squares of all possible values of a1.
24. A positive integer is called
friendly if it is divisible by the sum of its digits. For example,
111 is friendly but 123 is not. Find the number of all two-digit friendly numbers.
25. In the diagram below, D and E lie on the side AB, and Flies on the side AC such that DA=DF=DE, BE= EF and BF= BC. It is given that L_ABC= 2 L_ACB. Find x, where L_BFD= X0•
B
F
26. In the diagram below, A and B(20,0) lie on the x-axis and C(O, 30) lies on they-axis such that L_ACB= 90°. A rectangle DEFG is inscribed in triangle ABC. Given that the area of triangle CGF is 351, calculate the area of the rectangle DEFG. y
c
A
D
0
6
E
B
X
27. Let
ABCDEF
be a regular hexagon. Let
G
be a point on
area of AGEF is 100, find the area of the hexagon G
E
ED
such that
ABCDEF.
EG =
3GD . If the
D
F
c B
A
28. Given a package containing 200 red marbles, 300 blue marbles and 400 green marbles. At each occasion, you are allowed to withdraw at most one red marble, at most two blue marbles and a total of at most five marbles out of the package. Find the minimal number of withdrawals required to withdraw all the marbles from the package.
29. 3 red marbles, 4 blue marbles and
5 green marbles are distributed to 12 students.
Each
student gets one and only one marble. In how many ways can the marbles be distributed so that Jamy and Jaren get the same colour and Jason gets a green marble?
30 . A round cake is cut into
n
pieces with 3 cuts. Find the product of all possible values of n.
31 . How many triples of non-negative integers ( x, y, z) satisfying the equation
xyz + x y + yz + zx + x + y + z = 2012? 32. There are 2012 students in a secondary school. Every student writes a new year card. The cards are mixed up and randomly distributed to students. Suppose each student gets one and only one card. Find the expected number of students who get back their own cards.
33. Two players
A
and B play rock-paper-scissors continuously until player
A
wins 2 consecutive
games. Suppose each player is equally likely to use each hand-sign in every game. What is the expected number of games they will play? 7
34. There are 2012 students standing in a circle; they are numbered 1, 2, . . . , 2012 clockwise. The counting starts from the first student ( number 1) and proceeds around the circle clockwise. Alternate students will be eliminated from the circle in the following way: The first student
stays in the circle while the second student leaves the circle. The third student stays while the fourth student leaves and so on. When the counting reaches number number
2012, it goes back to
1 and the elimination continues until the last student remains. What is the number
of the last student?
n
< n.
35. There are k people and chairs in a row, where 2 ::; k There is a couple among the k people. The number of ways in which all k people can be seated such that the couple is seated together is equal to the number of ways in which the (k - 2) people, without the couple present, can be seated. Find the smallest value of n.
8
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2012 Junior Section (First Round)
Multiple Choice Questions
1. Answer: (D) . Since x 2 + 2bx + (b - 1) 0, we have a/3 b - 1 and a + f3 -2b. Then (a- /3) 2 (a + /3) 2 - 4af3 ( -2b) 2 - 4(b - 1) 4(b2 - b + 1) 4[(b - 1 2) 2 + 3 4] 3 The equality holds if and only if b 1 2. 2. Answer: (E) . Note that n 2o 12 n 201o n 201o (n 2 + 1). If n 2, then 5 (n 2 + 1) and 2 n2010 . If n 13 or 47, then 10 (n 2 + 1). If n 35, then 2 (n 2 + 1) and 5 n 2010 . If n = 59, then 5 f (59 2 + 1) and 5f n 2010 . 3. Answer: (B). There are 8 vertices in a cube. Any 4 vertices form a triangular pyramid unless they lie on the same plane. (!)- 6 -6 58 4. Answer: (B). Let the radius of the circle be 1. Then AE CE 2 CE 2(1 - 1 J2) 2 _ J2 sl::.ABc AE cE 1 OD � AE OD Sl::.AOD Sl::.AOD =
=
=
=
=
=
=
�
=
+
=
=
=
=
=
x
x
x
=
x
9
x
=
=
c
D
5. Answer: (A) . DE . Then = EF . Note that Let AE EB DF + FC EF + FG EF + EB - 2EF + 2 1 - DC AB - AB AB - EB GB - EB EF - (I__ ) 3 - V5 (0 1). Then + - 3 = 0, and thus = -;1 2 6. Answer: (D) . x
x
=
_
_
_
_
x
< x<
x
x
_
Consider the extreme cases:
Then Xmin
= 8 - 24� = 4 - 2� and
7. Answer: (D)
By Heron's formula,
Then
Xmax
= �28+ 1 = 8(� - 1).
s6.ABC s6.ABD s6.ACD
=J12(12 - 5)(12 - 8)(12 - 11) 4v'2I. 5 � X AB X AD X sin a AB � X AC X AD X sin a AC 11 5v'2I 5 S6.ABD = 5 + 11 X 451 =-4 =
10
x
B
c
A 8. Answer: (B) .
� = b ( l�J + { �} ) = b l�J + b { �} · It follows from a = b l�J + b { �} that b l�J = a. Hence, � = l�J is an integer. Note that a = b ·
Obviously, b is not necessary an integer even if a is an integer.
9. Answer: (C) .
0, then = -= ( -x) = �x -= --2x = -2.
x can take any nonzero real number. If x
If x
<
0, then
y
=
-=
-fE- -fE
X
-fE-
X
-
-fE- -$
y
>
-fE
X
- xX
-8-
=- = X
-
X
X
10. Answer: (C) .
The assignment is as follows:
A: in Z, teaches Music; B: in
Y,
teaches Latin;
C:
in X , teaches Mathematics.
Short Questions
11. Answer: 4. 75 (2a + 2b )(2-a 2-b ) = 2 2a-b + 2b-a . Then 2a-b 2b-a = 417 = 4 4"1 12 Since a b, we obtain 2a-b = 4. 12. Answer: 25. x 3 - x + 120 120 . It 1s. an .mteger "f and on y . f ( x2 - 1) -120 . =x Note t h at (x 1) (x 1) x2 1 Then x2 - 1 = ±1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120. Hence, x = 0 2 3 4 5 11; and 2 + 3 + 4 + 5 + 11 = 25. 13. Answer: 9. +
=
+
+
+
>
_
+
+
1
_
11
1
1
0.
Let y =v'3x 2 - 8x +
1 . Then the equation becomes y y'3 2 - 11 = 3 . Theny'3y 2 - 11 = 3 - y. Squaring both sides, we have 3y 2 - 11 = 9 - 6y y 2 + 3y - 10 = 0 . Then y = 2 or y = -5 (rej ected because y 0) . Solve 3x 2 - 8x 1 = 2 2 . Then x = 3 and x = -3- 1 . Hence, k = 3/3- 1 = 9 . 14. Answer: 2996 . y
+
2
+
+
y 2 ; that is,
Rewrite the equation as the following:
b c d b c d 0 0 + c b a Since a is the last digit of 2d, a is even; since 2a 1::::; d::::; 9, a::::; 4 . So a = 2 or a = 4 . If a = 4, then d 2a 1 = 9 and thus d = 9; but then the last digit of 2d would be 8 1- a, a a a 1 d
2
+
2
+
0
+
contradiction.
a = 2, then d 2a 1 = 5 and the last digit of 2d is 2; so d = 6 . The equation reduces to b c b c 1 1 c b There are 2 cases : either 2c 1 = b and 2b = 10 c, which has no integer solution; or 2c + 1 = 10 b and 2b 1 = 10 + c, which gives b = c 9 . 15 . Answer: 110 . Complete the square: (x - 11) 2 + (y - 10) 2 = 0 . Then x = 11 and y = 10; thus xy = 110 . 16 . Answer: 93 . If
+
+
+
+
+
=
Rearranging, we have
- 876 8 mn2 - 217n = 4mn - 876 n = 4mn mn - 21 7 = 4 - mn - 217 Then ( mn - 217) 8 . It follows that mn - 217 = ±1, ±2, ±4, ±8 . So mn = 218, 216, 219, 215, 221, 213, 225, 209, and n = -4, 12, 0, 8, 2, 6, 3, 5 respectively. mn 216 225 N ot e th at m = - 1s a positiVe mteger. T en m = - = 18 or m = - = 75 . n 12 3 =?-
.
. .
.
h
12
17. Answer: 50. Write x = lxJ + --fl: � Then 100:::; (lxJ + --fl:-) 2 - lxJ 2 = 2lxj--fl: -+ --fl:...J, 2lxJ + 1. S o lxJ 50 and x 2 lx2J = 100 + 502 = 2600. On the other hand, x = v'26QO is a solution. 18. Answer: 35. (X! + 2X 2 + + 49X 49 ) 2 = ( 1 X! + J2 J2X 2 + " " " + .j49 " .j49X 49 r :::; ( 1 + 2 + + 49) (xi + 2x� + + 49x�9) = 49 X2 50 X 1 = 352 . = X 49 = ± 351 The equality holds if and only if x1 = 19. Answer: 58. �
<
�
·
·
·
·
·
·
·
·
·
·
·
·
·
·
·
As shown below,
J x2 + (20 -y) 2 + Jy 2 + (21- z) 2 + J z2 + (20 -w) 2 + Jw 2 + (21 - x) 2 = AB + BC + CD +DA = A'B + BCJ + CD + DA." :::; A' A" = 422 + 402 = 58. 21-x A" A' A x • ::: ------- -.----"'7""o;::,...-----.,.. ----------:::,.. ', ;::,:, ,.. .... ....... """" ,' ;3 , .... ..I .... ',
,..,
o C'l
;3 I
B
'
21- z
C\
\ \
' \
\ jj
..
...
........
,,
...........
.......
,, A.. -" '
20. Answer: 1818.
A = abed. Then 1000(a + n) + 100(b - n) + 10(c + n) + (d - n) = nA. It gives A + 909n = nA; or equivalently, ( n - 1 )A = 909n. Note that (n - 1) and n are relatively prime and 101 is a prime number. We must have (n - 1) So n = 2 or n = 4. If n = 4, then A = 1212, which is impossible since b n. So n = 2 and A = 909 2 = 1818. Let the 4-digit number be
-9.
<
13
x
21. Answer: 4. Note that 1024 = 21 0 1 (mod 1023). Then 10211022 ( _ 2)1022 21022 21ox102+2 1024102 22 1102 22 4 22. Answer: 375. Let m and M be the smallest and the largest numbers. Then m + 80 + 1 = M + 80 1 = m + M + 80 5 5 6 Solving the system, m = 15 and M = 25. Then mM = 375. =
=
=
=
X
=
=
X
=
(mod
1023)
_
23. Answer: 3 .
+
1 a3 =- 1-,a a4 =--2 +- a, as =-3--, + 2a . . . . In general, a1 = a. Then a2 =-1 -, + a 2 + a 3 +2 a 5 + 3a an =FFnn+l++FnF-na1 a where F1 = 0, F2 = 1 and Fn +l = Fn Fn-1 for all 1. F + F2o11 a = a, then (a2 + a - 1)F2o12 = 0. If a 201 2 = 2o1 2 + L' 2012 a 2 L' 1 20 3 Since F 201 2 0, we have a + a - 1 = 0. Let and (3 be the roots of a 2 + a - 1 = 0. Then Let
-=-----'-==----
+
v
n�
v
a
>
23. ga + 1, we have (a + b) ga. . lOa + b = -Smce a+b a+b Case 1: If 3 f (a + b), then (a + b) a, and thus b = 0. We have 10 20 40 50 70 80. 3a ::::; 3. If 3a = 1, then Case 2: If 3 (a + b) but gf(a + b), then (a + b) 3a and 1::::; a+b a3a+ b 3a 2a = b, we have 12 24 48. If a + b = 2, then a = 2b, we have 21 42 84. If -= 3, then a + b b = 0, we have 30 60. Case 3 : If g (a+ b), then a+b = g or 18. If a+b = 18, then a = b = g, which is impossible. If a + b = g, then we have g friendly numbers 18 27 36 45 54 63 72 81 go.
24. Answer:
--
Therefore, in total there are 23 2-digit friendly numbers.
108. Since DA = DE = DF, LEFA = goo. Let LEBF = LEFB = x0• LBFC =goo - x0 and LCBF = 2x0, LBAC =goo - 2x0•
25. Answer:
14
Then
LBCF =
B
F
It is given that 3x = 2 ( 9 0 - x ) . Then x = 36. So x = 180 - ( 90 - x ) - ( 90 - 2x ) = 3x = 108. 26. Answer: 468.
AO = 30202 = 45. Then the area of !::,. ABC IS. ( 20 + 452 ) Let the height of !::,. CGF be Then 3 ( ) 2 = 351 = ( 3 ) 2 =
Note that
X
30
= 975 .
h.
h
30
Note that the rectangle
5
975
:::?
30
h
- h
DEFG has the same base as !::,. CGF. Then its area is 351
X
2 3
X
2 = 468
27. Answer: 240.
DE = 1 and denote A = EG = 3 4. EGAX = 1 +A AGEF 1 +A 1 1 + 2A . - -8 - Imp Ies 4 4 DEXY DEXY 8 ABCDEF 6 AGEF 1 + 2A Note that DEXY = 8" Then ABCDEF 6 12 If AGEF = 100, then ABCDEF = 100 = 240. S E A G D We may assume that -
2
1.
-
x
X/
I
I
I I
I
Fl
I
I
I
I
.'-------------.lL.----->L
A
B
28. Answer: 200. 15
5 12
Since at most one red marble can be withdrawn each time, it requires at least
200 withdrawals.
200 is possible. For example, 150 . (1, 2, 2) + 20 . (1, 0, 4) + 10 . (1, 0, 2) + 20 . (1, 0, 0) (200, 300, 400). 29. Answer: 3150. Case 1: Jamy and Jaren both take red marbles. So 1 red, 4 blue and 4 green marbles are distributed to 9 students: (�) (�) 630. Case 2: Jamy and Jaren both take blue marbles. So 3 red, 2 blue and 4 green marbles are distributed to 9 students: G) (�) 1260. Case 3: Jamy and Jaren both take green marbles. So 3 red, 4 blue and 2 green marbles are distributed to 9 students. The number is the same as case 2. 630 + 1260 + 1260 3150. 30. Answer: 840. With three cuts, a round cake can be cut into at least 1 + 3 4 pieces, and at most 1 + 1 + 2 + 3 7 pieces. Moreover, 4, 5, 6, 7 are all possible. 4 5 6 7 840. On the other hand,
=
=
X
=
X
=
=
=
n=
x
x
x
=
31. Answer: 27. ( + 1)(y + 1)( z + 1) 2013 3 11 61. I f all y, z are positive, there are 3 ! 6 solutions. If exactly one of y, z is 0, there are 3 6 18 solutions. If exactly two of y, z are 0, there are 3 solutions. 6 + 18 + 3 27. 32. Answer: 1. 1 . Then the expectation For each student , the probability that he gets back his card is 20 12 1 . of the whole class 2012 2012 1. x
=
x,
=
x
=
x,
=
x
x,
IS
x
x
=
=
16
33. Answer: 12. Let E be the expectation. If
A
does not win, the probability is
If A wins and then does not win, the probability is
2/3 and the game restarts.
(1/3)(2/3) and the game restarts. The probability that wins two consecutive games is (1/3)(1/3). Then E = 32 (E + 1) + 92 (E + 2) + 91 2. Solving the equation, we get E = 12. 34. Answer: 1976. If there are 1024 = 21 0 students, then the 1024th student is the last one leaving the circle. Suppose 2012 - 1024 = 988 students have left. Among the remaining 1024 students, the last student is (2 988 - 1) + 1 = 1976. 35. Answer: 12. (n - 1) 2 (� = �) ( k - 2)! = (k : 2) ( k - 2)!. Then 2 = (n k + � n k + 2 ) That is, 2n 2 - (4k - 6)n + (2k 2 - 6k + 4 - n) = 0. We can solve 7 - 5 (n k). v8k --n = k +-4 Note that the square of any odd number has the form 8k-7. Choose k so that v8k - 7-5 = 4, i.e . , k 11. Then n 12. A
X
X
x
x
x
x
x
x
_
>
=
=
17
1
_
"
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2012
(Junior Section, Round 2) 0930-1230
Saturday, 23 June 2012
0 be the centre of a parallelogram ABCD and P be any point in the plane. Let M, N be the midpoints of AP, BP, respectively and Q be the intersection of MC and ND. Prove that 0, P and Q are collinear.
1. Let
A such that each of the ten digits 0, 1, exactly once as a digit in exactly one of the numbers A, A2, A3.
2. Does there exist an integer
. . . , 9 appears
L.ABC, the external bisectors of LA and LB meet at a point D. Prove that the circumcentre of L.ABD and the points C, D lie on the same straight line.
3. In 4.
Determine the values of the positive integer n for which the following system of equations has a solution in positive integers x1, x2, , Xn. Find all solutions for each such n. •
.
Xl+ X2 + · · · + Xn = 16 1 1 1 + + . . ·+ - = 1 Xn X1 X2 5.
.
(1) (2)
Suppose S = a1, a2, , a15 is a set of 1 5 distinct positive integers chosen from 2 , 3, ... , 2012 such that every two of them are coprime. Prove that S contains a prime number. (Note" Two positive integers m, n are coprime if their only common factor is 1.) •
.
.
18
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2012
(Junior Section, Round 2 solutions)
MN II AB II CD, we have �MQN"' �CDQ. Hence MN = AB/2 = CD/2. Thus QM = CQ/2. In �ACP, CM is a median and Q divides CM in the ratio 1:2. Thus Q is the centroid. Hence the median PO passes through Q.
1. Since
p
Since the total number of digits in A, A2 and A3 is 10, the total number digits in 32, 322, 32 3 is 11 and the total number of digits in 20, 202, 203 is any solution A must satisfy 21 :::; A :::; 31. Since the unit digits of A, A2, A3 are distinct, the unit digit of A can only be 2, 3, 7, 8. Thus the only possible values of A are 22, 23, 27, 28. None of them has the desired property. Thus no such number exists.
2.
9,
CD bisects LC. If CA = CB, then CD is the perpendicular bisector of AB. Thus the circumcentre of �ABD is on CD. 3. Note that
If C A =/:. CB, we may assume that CA > CB. Let E be a point on CB extended and F be the point on CA so that C F = CB. Then, since CD is the perpendicular bisector of
19
BF, we have LAFD = LDBE = LDBA. Thus AFBD is a cyclic quadrilateral, i.e . , F is on the circumcircle of 6.ABD. The circumcentre lies on the perpendicular bisector of BF which is CD. 4.
then Without loss of generality, we may assume that x1 S x2 S S Xn· If x 1 = from n = and cannot be satisfied. Thus x1 � If x2 = then n = 2 and again cannot be satisfied. Thus x2 � Similarly, X3 � Thus x4+ + Xn S with x4 � Thus n s ·
(2), (1)
1
(1) 4.
4.
·
2.
3.
1,
·
4.
2,
·
·
7
·
n =
(i)
1: No solution. (ii) 2: The only solution of + 1 is x1 x2 2 which doesn't satisfy (1) . Thus there is no solution. (iii) 3: The only solutions of; + +;3 1 are (x1 , x2 , x3) (2, 3, 6) , (2, 4, 4) and ( 3, 3, 3) . They all do not satisfy (1) . (iv) 4: According to the discussion in the first paragraph, the solutions of x 1 + + X4 16 are (x1 , x2 , x3 , x4) (2, 3, 4, 7), (2, 3,5, 6) ,(2, 4,4, 6) ,(2, 4,5,5), (3, 3, 4, 6) , (3, 3,5,5),(3, 4,4,5),(4, 4, 4,4) . Only the last one satisfy (2) . Thus the system of equations has a solution only when 4 and for this the only solution is x1 x2 X3 X4 4. __!_
n =
Xl
n =
n =
·
·
·
1
__!_ 2 X
x12
=
=
=
=
=
=
=
=
=
=
n =
=
n,
5.
Suppose, on the contrary, that S contains no primes. For each i, let Pi be the smallest prime divisor of ai . Then Pl, P2 , . . . , P1 5 are distinct since the numbers in S are pairwise coprime. The first primes are 5, If = = > and aj � Pj is the largest among Pl, P2 , . . . , p1 5 , then Pj � a contradiction. Thus S must contain a prime number.
15
2, 3, 7, 11, 13, 17, 19 , 23, 29 , 31, 37, 41, 43, 47. 47 472 472 2309 2012,
20
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO)
2012
Senior Section (First Round)
Tuesday,
29
May
2012
0930-1200
hrs
Instructions to contestants 1.
2.
3.
Answer ALL
35
questions.
Enter your answers on the answer sheet provided. For the multiple choice questions, enter your answer on the answer sheet by shading the bubble containing the letter (A, B, C, D or E) corresponding to the correct answer.
4- For the other short questions, write your answer on the answer sheet and shade the appropriate bubble below your answer. 5. 6. 7.
8.
No steps are needed to justify your answers. Each question carries
1
mark.
No calculators are allowed. Throughout this paper, let LxJ denote the greatest integer less than or equal to x. For example, L2.1J = 2, L3.9J = 3.
PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO
21
Multiple Choice Questions
1. Suppose
a
and (3 are real numbers that satisfy the equation
) V J2+ 1
(V
x2 + 2 J2+ 1 x+
�3 +;3 .
=
0
Find the value of
(A) 3 J J2+ 1(J2 - 1) - 8
(B) 8 -6v\12 + 1(J2 - 1)
(C) 3 JJ2+ 1(J2 - 1)+ 8
(D) 6 J J2+ 1(J2 - 1) - 8
(E) None of the above
2. Find the value of 20112 X 2012 - 2013 20132 X 2014 - 2015 -- -------+ -------,-2012! 2014! 1 + 1 - 1 - 1 2 0 1 0! 2 011! 2 01 2 ! 2 009! 1 + 13 + 1 (D) 1 + 2 010! 2 01 ! 2 01 4 ! 2 009!
1 + 1 + 1 + 1 2 010! 2 011! 2 01 2 ! 2 009! 1 + 1 - 1 - 13 (C) 2 010! 2 01 2 ! 2 01 ! 2 009! 1 + 1 - 13 - 1 (E) 2 0 1 0! 2 01 ! 2 01 4! 2 009! (A)
(B)
3. The increasing sequence T = 2 3 5 6 7 8 10 11 consists of all positive integers which are not perfect squares. What is the 2012th term of T? (A) 2055
(B) 2056
(C) 2057
(D) 2058
(E) 2059
4. Let 0 be the center of the inscribed circle of triangle �ABC and D be the point on AC with OD _LAC. If AB = 10 AC = 9 BC = 11, find CD.
(A) 4
(B) 4.5
(C) 5
(D) 5.5 22
(E) 6
5. Find the value of
(A)
1
cos4 75° + sin4 75° + 3 sin2 75° cos2 75° cos6 75° + sin6 75° + 4 sin2 75° cos2 75° ·
(B) �
(C)
�
(D) 1
(E) cos 75° + sin 75°
6. If the roots of the equation x2 + 3x - 1 = 0 are also the roots of the equation x4 + ax2 + bx+ c = 0, find the value of a+ b+ 4c. (A) -13
(B) -7
(C) 5
(D) 7
(E) 11
7. Find the sum of the digits of all numbers in the sequence 1, 2, 3, 4, . . . , 1000. (A) 4501
(B) 12195
(C) 13501
(D) 499500
(E) None of the above
8. Find the number of real solutions to the equation X
100
=
. smx,
where x is measured in radians. (A) 30
(B) 32
(C) 62
(D) 63
(E) 64
9. In the triangle 6ABC, AB = AC, LABC = 40° and the point D is on AC such that BD is the angle bisector of LABC. If BD is extended to the point E such that DE= AD , find LECA. A
(A) 20°
(B) 30°
(C) 40°
(E) 50°
10. Let m and n be positive integers such that m > n. If the last three digits of 2012m and 2012n are identical, find the smallest possible value of m+ n. (A) 98
(B) 100
(C) 102
(D) 104
23
(E) None of the above
Short Questions
11. Let a b c and d be four distinct positive real numbers that satisfy the equations (a2m2 c2 o12)(a2012 d2012) 2011 and (b2o12 c2 o12) (b2o12 d2o12) 2011 Find the value of (cd ) 2 01 2 -(ab ) 2 012 . 12. Determine the total number of pairs of integers x and y that satisfy the equation 1 1 1 +2 3' 2X _
_
_
=
_
Y
=
Y
13. Given a set S 2 , a collectionF of subsets of S is said to be intersecting if for any two subsets A and B in F, we have A n B -=1 0. What is the maximum size ofF? =
1
10
14. The set M contains all the integral values of m such that the polynomial 2 ( m -1) x2 - ( m2 - m +12)x + 6m has either one repeated or two distinct integral roots. Find the number of elements of M.
15. Find the minimum value of
I
sin x+ cos x
+ coscosx -2xsin x I ----
16. Find the number of ways to arrange the letters A, A, B, B, C, C, D and E in a line, such that there are no consecutive identical letters. x
17. Suppose x 3V2+Iog3 is an integer. Determine the value of x. 18. Let f(x) be the polynomial (x-a1 )(x-a2)(x-a3)(x-a4)(x-as) where a1 , a2, a3, a4 and a5 are distinct integers. Given that f (104) 2012, evaluate a1+a2+a3+a4+a5. 19. Suppose x, y, z and are positive real numbers such that yz 6-Ax xz 6-Ay xy 6-Az x2+y2+z2 1 Find the value of (xyz.A) -1 . =
=
A
=
=
=
=
24
20. Find the least value of the expression real numbers satisfying the equation
(x +y )(y +z) , given that x, y, z are positive xyz(x+y+z) 1. =
21. For each real number let be the minimum of the numbers 2012. 4. Determine the maximum value of -2
x+
x, f(x)
6f (x) +
4x+1, x + 2 and
22. Find the number of pairs (A, B) of distinct subsets of 1, 2, 3, 4, 5, 6 such that A is a proper subset of B. Note that A can be an empty set. 23. Find the sum of all the integral values of that satisfy
x
JX+3- 4 �+ JX+ 8- 6� 24. Given that for real values of
=
1.
I x2+ 4x +5- v'x2+ 2x+5 1 ,
S = v'
x, find the maximum value of 84.
25. Three integers are selected from the set S = 1, 2, 3, ... , 19, 20 . Find the number of selections where the sum of the three integers is divisible by 3. 26. In the diagram below, ABCD is a cyclic quadrilateral with AB = AC. The line FG is tangent to the circle at the point C, and is parallel to BD. If AB = and BC = 4, find the value of 3AE.
6
A
25
27. Two Wei Qi teams, A and B, each comprising of 7 members, take on each other in a competition. The players on each team are fielded in a fixed sequence. The first game is played by the first player _of each team. The losing player is eliminated while the winning player stays on to play with the next player of the opposing team. This continues until one team is completely eliminated and the surviving team emerges as the final winner - thus, yielding a possible gaming outcome. Find the total number of possible gaming outcomes. 28. Given that m ( cos + ( sin and n ( .J2- sin ( cos where and are the usual unit vectors along the x-axis and the y-axis respectively, and E ( 1r 2n ) . If the length or magnitude of the vector m n is given by m n 8'f, find the value of 5 cos (� + i) 5.
= B) i
B) j
=
+
B) i + B ) j , i j B + =
+
29. Given that the real numbers x, y and the maximum value of j ( x y v2 x
z satisfies the condition x +y + z = 3, find +13 +�3y +5+V'8z +12. 30. Let P ( x ) be a polynomial of degree 34 such that P(k) = k (k + 1 ) for all integers z) =
k=0 1 2
34. Evaluate 42840 x P ( 35) .
31. Given that a is an acute angle satisfying
v369 - 360 cosa +v544 - 480 sina - 25 = 0 find the value of 40 tana. 32. Given that and
a b c d e are real numbers such that a+b +c+d+e = 8 a2+b2+ c2+d2+e2 = 16
Determine the maximum value of L eJ. 33. Let L denote the minimum value of the quotient of a 3-digit number formed by three distinct digits divided by the sum of its digits. Determine L10 £J . 34. Find the last 2 digits of
35. Let
f (n) be the integer nearest to yfii. Find the value of
26
Singapore Mathematical Society
Singapore Mathematical Olympiad (SMO) 2012 Senior Section (First Round Solutions)
Multiple Choice Questions
1. Answer:
(D)
+
Note that a (3 =
-
(2V J2 +1) and a(3 J J2 +1. __!_ __!_ (a+ (3) 3- 3a(3(a + (3) + =
a3
(33
=
=
( af3) 3 s..)J2
+1 6vv'2+1 6(vv'2 +1)- s(J2 +1) v'2+1 6v,---.J'i- +-1 ( .J2 1) s -
2. Answer: (E)
27
-
3. Answer: ( C ) Note that 442 = 1936, 452 = 2025 and 462 = 2116. So 2, 3, . . . , 2012 has at most 2012 - 44 terms. For the 2012th term, we need to add the 44 numbers from 2013 to 2056. But in doing so, we are counting 452 = 2025, so the 2012th term should be 2012 + 44 + 1= 2057. 4. Answer: ( C )
B
is tangent to the circle at D, by constructing E and F as shown, we have CD= CF, AD= AE and BE= BF. Solving for the unknowns give CD= 5. AC
5. Answer: (D) cos6 x + sin6 x+ 4 sin2 x cos 2 x = ( cos 2 x + sin 2 x ) ( sin4 x + cos 4 x - sin 2 x cos 2 x ) + 4 sin 2 x cos 2 x = ( sin4 x+ cos4 x + 3 sin2 x cos 2 x ) . 6. Answer: ( B ) We have the factorization ( x2 + 3x - 1 ) ( x2 + mx - c) = x4 + ax 2 + bx + c.
Comparing coefficients give 3 + m= 0, -1 - c + 3m= a and -3c - m= b . We can solve these equations to obtain a+ b + 4c= -7. 7. Answer: ( C ) Among all numbers with 3 or less digits, each i, i= 0, 1, 2, , 9, appears exactly 300 times. Thus the sum of the digits of all the numbers in the sequence 1, 2, 3, 4, , 999 is 300 ( 1 + 2 + . . . + 9 ) = 13500, and so the answer is 13501. ·
·
·
·
28
·
·
8. Answer: (D) Since -1:::;; sin x:::;; 1, we have - 100:::;; x:::;; 100. We also observe that 317r < 100 < 327r. For each integer k with 1 :::;; k :::;; 16, 1 �0 = sin x has exactly two solutions in [2(k-1)7r (2k - 1)7r], but it has no solutions in ((2k - 1)7r 2k7r). Thus this equation has exactly 32 non-negative real solutions, i.e. x = 0 and exactly 31 positive real solutions. Then it also has exactly 31 negative real solutions, giving a total of 63. 9. Answer: ( C ) Construct a point F on EC such that EF= EA. Since LAED LFED, 6.AED is congruent to 6.FED. Thus DF = DA = DE and LFDE = LADE = 60°. We also have LEDC = LADE= 60°, which implies that LFDC= 60° and 6.CFD is congruent to 6.CED. In conclusion, LEGD= LFCD= 40°. =
A
E
10. Answer: (D) We want to solve 2012m
_
2012n (mod 1000) which is equivalent to
Since (12m-n - 1) is odd, we must have 8 12n so n 2:: 2. It remains to check that 125 12m-n - 1 i.e. 12m-n 1 ( mod 125). Let
29
Short Questions
11. Answer: 2011 Let A = a2 01 2 , B = b2 01 2 , C = c2 0 1 2 and D = d2 01 2 . Then A and B are distinct roots of the equation (x -C ) (x - D) = 2011. Thus the product of roots, AB = CD - 2011 and CD - AB = 2011. 12. Answer: 6 Note that x and y must satisfy 2x+ 1 · 3 = y(y + 2). We first assume x :;::: 0, which means both y and y + 2 are even integers. Either 3 y or 3 y + 2. In the first case, assuming y > 0, we have y = 3 · 2k and y + 2 = 2(3 . 2k-l + 1) = 2x+1 -k. The only way for this equation to hold is k = 1 and x = 3. So (x, y, y + 2) = (3, 6, 8). In the case 3 y+ 2, assuming y > 0, we have y+ 2 = 3· 2k and y = 2(3· 2k-1 - 1) = 2x+ 1 -k. Now the only possibility is k = 1 and x = 2, so (x, y, y + 2) = (2, 4, 6) . In the two previous cases, we could also have both y and y+ 2 to be negative, giving (x, y, y + = (3, -8, -6) or (2, -6, -4). Finally, we consider x < 0 so 3 = 2-x-ly(y + 2). In this case we can only have x = -1 and (x, y, y + 2) = (-1, 1, 3) or (-1, -3, -1). Hence possible (x, y) pairs are (3, 6) , (3, -8) , (2, 4) , (2, -6) , ( -1, 1) and ( -1, -3).
2)
13. Answer: 512 If A E F then the complement can belong to F, that is
S nA
fj. F. So at most half of all the subsets of
F:::;
S
210
2 = 512.
Equality holds because we can take F to be all subsets of S containing 1. 14. Answer: 4 If m = 1, the polynomial reduces to -12x + 6 = 0 which has no integral roots. For m-=/:- 1, the polynomial factorizes as ((2x - m) ( (m - 1)x - 6) , with roots x = r; and x = � . For integral roots, m must be even and m - 1 must divide 6. The m l values are m = -2, 0, 2 and 4. So M has 4 elements. only possible 15. Answer: 2 Note that cos 2x = cos2 x - sin2 x. So 1 cos x - sin x = sin x + cos x + . sin x + cos x + smx + cos x cos 2x Set w = sin x + cos x and minimize lw + �I· By AM-GM inequality, if w is positive then the minimum of w + � is 2; if w is negative, then the maximum of w + � is -2. Therefore, the minimum of lw + �I is 2.
I
I I
30
I·
16. Answer: 2220 We shall use the Principle of Inclusion and Exclusion. There are ways to arrange the letters without restriction. There are ways to arrange the letters such that both the As occur consecutively. (Note that this is the same number if we use B or C instead of A.) There are � ways to arrange the letters such that both As and Bs are consecutive. (Again this number is the same for other possible pairs.) Finally there are 5! ways to arrange the letters such that As, Bs and Cs occur consecutively. For there to be no consecutive identical letters, total number of ways is
21�i21
2i�1
� - 3 X _2!_ + 3 X 212121
2121
17. Answer: 9 J2 log3 Taking logarithm, we get log3 J2 solution for is 2. Therefore
y = +y
'
6! - 5. - 2220
21
-
x = + 2x Let y = log3 x. x = 3 = 9.
The only possible
18. Answer: 17 The prime factorization of 2012 is 2 503. Let b 104. If ai are distinct, so are b- ai, i.e. (b- a1 ) , (b- a3 ) , (b- a4 ) and (b- a5 ) must be exactly the integers 1 -1 2 -2 503 . Summing up, we have
= b( -a2) , 2 5(104) - ( a1 + a2+ a3 + a4 + a5 ) = 1 - 1 + 2 - 2 + 503 ·
19. Answer: 54 Multiplying the first three equations by
x, y and z respectively, we have xyz = 6-\x2 = 6-\y2 = 6-\z2 Since,\# 0 and x2+y2+z2 = 1, we deduce that x2 = y2 = z2 = !,sox = y = z = }s and x2 1 2 Hence \
_
/\-
xyz 6 x2
_
-
x3
- 6V3" _
20. Answer: 2 Observe that 1 (x+y)(y +z) = xy +xz+y2+yz = y (x+y +z) +xz = -+ xz xz 2: 2 where the equality holds if and only if xz = 1. Let x = z = 1 and y = .J2- 1, then we have the minimum value 2 for (x + y )(y + z ) . 31
21. Answer: 2028 Let L1, L2, L3 represent the three lines y + 1, y + 2 and y + respectively. Observe that L1 and L2 intersects at G, D, L1 and L3 intersects at (�, 3), and L2 and L3 intersects at Thus
= 4x
=x
=-2x 4
(�, �) .
1.
X< x=r3 ' �f (x) = x 2, 3 <X< X=�-�' 3 ' -2x 4, X> Thus the maximum value of f (x ) is � and the maximum value of 6f (x ) 4x+ 1, 7
3' 8
3'
1
+
3·
+
+
2028.
2012 is
22. Answer: Since B cannot be empty, the number of elements in B is between 1 to After picking B with k elements, there are 2k - 1 possible subsets of B which qualifies for A, as A and B must be distinct. Thus the total number of possibilities is
665
6.
t m (2.
_
1)
=t, m (2. 1) = t, m 2· - t, m =(2 1) 6 - (1 1) 6 =36 - 26 =665. _
+
+
23. Answer: The equation can be rewritten as (-,fx� - 2) 2 + Cv'x=-I. - 3)2 1. If -yX=l 2: 3, it reduces to -yX=l - + -yX=l - 3 1 i.e. -yX=l 3 giving 10. If -vX=1 ::;; 2, it reduces to 2 - -yX=l + 3 - -yX=l 1 i.e. -yX=l 2 giving
45
J
J
= 2 = = X= = = X=5. If 2 <-yX=l<3, i.e. 5<x <10, it reduces to -v'X=l - 2 3 - -yX=l = 1 which is true for all values of x between 5 and 10. Hence the sum of all integral solutions is 5 6 7 8 9 10=45. 24. Answer: 4 2) 2 (0 - 1)2 -y'(x 1) 2 (0 - 2)2 1 . =y'(x I +
+
S
+
+
+
+
+
32
+
+
+
Let P = 0) , A = and B = then S represents the difference between the lengths PA and PB. S is maximum when the points P, A and B are collinear and that occurs when P = 0). So
(x,
(-2, 1)
(-1, 2) ,
(-3,
Thus the maximum value of S 4 =
4.
25. Answer: 384 Partition S into three subsets according to their residues modulo 3: S0 = 3, 6, . . . , 18 , = 1, 4, . . . , 19 and = 2, 5, . . . , 20 . In order for the sum of three integers to be divisible by 3, either all three must belong to exactly one or all three must belong to different Hence total number of such choices is (�) + 2 G) + 6 7 7 = 384. 26. Answer: 10 82
81
Si
si.
x
x
Since BD II FG and FG is tangent to the circle at C, we have LBCF = LOBE = LDCG = LBDC = LBAC. Furthermore LBEC = LBAC+ LABE = LOBE+ LABE = LABC = LACE. We can then conclude that BE = BC = DC = t:.DCE. If we let AE =
4. Also, f:.ABE is similar to DE AE 2 ===> DE = 3x. = DC AB By the Intersecting Chord Theorem, AE · EC = BE· ED, i.e. x(6- x ) = 4(�x ) , which gives x = so 3AE = 3x = 10. 27. Answer: 3432 x,
1 3°,
We use a1, a2, , a7 and b1, b2, , b7 to denote the players of Team A and Team B, respectively. A possible gaming outcome can be represented by a linear sequence of the above terms. For instance, we may have a1a2b1b2a3b3b4b5a4b5b7a5a6a7 which indicates player followed by player from Team A were eliminated first, and the third player eliminated was player from team B. However Team A emerged the final winner as all seven players of Team B gets eliminated with a5, a6 and a7 remaining uneliminated. ( Note a6 and a7 never actually played. ) Thus, a gaming outcome can be formed by choosing out of possible positions for Team A, with the remaining filled by Team B players. Therefore, the total number of gaming outcomes is given by C 74) = c�D'2 = •
14
•
•
•
1
•
•
1
2
7 14 3432.
33
28. Answer: 1
8� =
m
=V4 + 2 J2 cosB - 2 J2 sinB = j2 1+ cos ( �) = 2 J2 I cos (� + i) I
+n
() +
Since 1r < () < 27r, we have �7f < � + i < �7f. Thus, cos (� + i) < that cos (� + i) -� and hence
=
29. Answer: Using the
f(x
0. This implies
5 cos (� + i) + 5 = 1
8
AM-GM
inequality, we have
y z) =v2x + 13+{!3y + 5+�8z + 12 =v 2x ; 13. V4+�3y: 5 . {12. {12+«8z; 12 . � . � . � 2x+13 + 4 3y+5 + 2 + 2 8z+12 + 2 + 2 + 2 < + + ----8--"' 2 4 3 1 29 = 4 (x+ y+ z)+ 4 4
4
=8
The equality is achieved at x
_____
= �, y = 1 and z = �.
30. Answer: Let n 34 and Q(x) (x+ 1)P(x)-x Then Q(x) is a polynomial of degree n + 1 and Q(k) Thus there is a constant C such that
40460
=
=
= 0 for all k = 0 1 2
Q(x) (x+ 1)P(x)-x Cx(x- 1)(x- 2 ) (x- n ) Letting x -1 gives 1 C(-1)( -2 ) ( -n - 1) C(-1)n+l (n+ 1)! Thus C (-1)n+l ( n+ 1)! and 1_ (x+Q(x))__1_ x+ (-1)n+lx(x- 1)(x- 2 ) (x- n ) P(x)__ - x+ 1 x+ 1 ( n+ 1)1.
=
=
·
·
·
·
·
·
=
=
=
So
·
·
=
·
(
·
·
·
)
1 n (-1)n+1(n+1)1 = 1 n+1+(-1t+l = -P(n+1) = -( n+ 1+ ) ) --( " n+ 2 (n+ l ) ! n+ 2 n+ 2 34
n.
since n
= 34 is even. Hence 42840 X P(35) = 34 X 35 X 36 X 3436=40460
31. Answer: 30 v'369 - 360 cos a and Let
a.
Y = v'544- 480 sin Observe that X2 = 122+152-2(12)(15) cos Y2 = 122+ 202-2(12)(20) cos (90°- ) and 152+ 202 = 252 = (X+ Y) 2, so we can construct a right-angled triangle ABD X=
a
a
as shown.
A
X
c y
15 2
B"---D ---=--____, In particular LABC a and LCBD fact similar to 6ABD. So LAD B a and
=
a.
= 90°- We can check that 6ACB is in = 40tana = 40 1520 = 30 x
32. Answer: We shall apply the following inequality:
3
4(a2+b2+c2+d2) � (a+b+c+d) 2 Since a+b+ c +d = 8-e and a2+ b2+ c2+ d2 = 16-e2, we have 4(16-e2) � (8-e) 2 i.e. e(5e -16) :::; 0. Thus 0 :::; e :::; 16 5. Notethatifa = b = c = d = 6 5,wehavee = 16 5. Hence LeJ = 3. 33. Answer: 105 A three-digit number can be expressed as lOOa+lOb+c, and so we are minimizing +lOb+ c F(a b c) = lOOaa+b+c 35
Observe that with distinct digits Thus we assume that Note that
a b c, F(a b c) has the minimum value when a< b
_
where the last congruence can be calculated by the extended Euclidean algorithm. Thus by repeated squaring, we have
19a 1933 59 _
(mod
100)
35. Answer: Note that (n+ � ) 2 n2 � ' so f(n2 n but f(n2 So each of the sequences (n - f(n) )�=1 ) and (n f(n))�=1 ) increases by for every increment of n by except when n If n 2 and (n we have n - f(n) so the former sequence has f(n every perfect square repeated once. On the other hand, if n we have n f(n) so the latter sequence but (n f(n omits every perfect square. Thus
5
= +n+
+n+1 = n+1. +n = = (0 1 1 2 ) + 2 ) = (2 3 5 26 1 1, = m +m. = m +m, =m + 1- + 1)) = m2, 2+ m, = m = m2+2m + 1+ + 1)) = m2+2m+ 2,
+
oo
�
G) f(n) + (�) -f(n) (�r
( �) + (�) � 3 � ( 3) () () -I: 3 + I: 3 + I: 3 - I: ( 3) oo
n-f(n)
oo
n+ f(n)
_
-
oo
_
2 n
00
m=1
n=O
=5
36
2 2 m
oo
n=1
2 n
oo
m=1
2 2 m
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2012
(Senior Section, Round 2) 0900-1300
Saturday, 23 June 2012
1. A circle
w
through the incentre I of a triangle ABC and tangent to AB at A, intersects the segment BC at and the extension of BC at Prove that the line IC intersects w at a point M such that
D
E.
MD=ME . 2. Determine all positive integers n such that n equals the square of the sum of the digits of n. 3. If 46 squares are colored red in a 9 9 board, show that there is a 2 2 block on x
x
the board in which at least 3 of the squares are colored red.
an an+l be a finite sequence of real numbers satisfying ao an+l and ak-1- 2ak ak+l
= =0 + Prove that fork = 0 1 n+1, ak 5.
Prove that for any real numbers a
<
=:;
1
for
k=1 2
n
c
=
k(n + 1-k) 2
c
b d 2:: 0 with a + b= +d
37
2,
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2012
(Senior Section, Round 2 solutions)
Join AD, ID, IA and AE. Let IE intersect AC at N. We have LIAN=LIAB= LIEA so that the triangles NIA and AlE are similar. Thus LAN!=LEAl=LIDB. Also LDCI = LNG!. Therefore, the triangles DC! and NCI are congruent. Hence LDIC=LNIC implying MD=ME.
1.
M
Let s ( n ) denote the sum of all the digits of n. Suppose n is a positive integer such that s ( n ) 2 = n. Let s ( n) =k so that n =k2 . Then s (k2 ) = s ( n ) =k. Let 10r-1 � k < lOr, where r is a positive integer. That is k has exactly r digits. From 10r-1 � k, we have r � logk + 1. From k < 10r, we have k2 < 102r so that k2 has at most 2r digits. Therefore, s (k2 ) � x 2r =18r � 18 logk + 18 which is less than k if k 2 50. Thus the equation s (k2 ) =k has no solution ink ifk 2 50. Let k < 50 and s (k2 ) =k. Taking mod we get k2 k ( mod Thus k- 0, 1 ( mod That is k =1, 10, 18, 19, 27, 28, 36, 37, 45, 46. Only when k =1 and k = we have s (k2 ) =k. The corresponding solutions for n are n=1 or 81. 2.
9
9) .
9,
9,
3. Suppose that at most 2 squares are colored red in any 2
9).
x
9,
2 square. Then in any x 2 block, there are at most 10 red squares. Moreover, if there are 10 red squares, then there must be 5 in each row.
9
38
Now let the number of red squares in row i of the 9 x 9 board be ri. Then ri+ri+1 ::; 10, 1:::; i:::; 8. Suppose that some ri :::; 5 with i odd. Then (r1 +r2)+· · ·+(ri-2+ri-1)+ri+· · · +(rs+rg):::; 4 x + 5 = 45. On the other hand, suppose that r1, r3, r5, r7, rg 2: 6. Then the sum of any 2 consecutive ri's is :::; 9. So (r1+r2)+· · ·+(r7+rs)+rg :::; 4 x 9+9 = 45.
10
=
k n 1- k Let bk = ( -+;, ). Then bo bn+l = and bk-1 - 2bk+ bk+l = -1 for k = 1, 2 . .. , n. Suppose there exists an index i such that ai rel="nofollow"> bi, then the sequence a0 b0, ..., an+1-bn+1 has a positive term. Let j be the index such that aj-1-bj-1 < aj-bj and aj- bj has the largest value. Then (aj-1- bj-1)+(aj+l- bj+1) < 2(aj- bj)· Using ak-1- 2ak+ak+1 2: -1 and bk-1- 2bk+bk+1 = -1 for all k we obtain (aj-1- bj-1)+(aj+1- bj+1) 2: 2(aj- bj) a contradiction. Thus ak:::; bk for all k. Similarly, we can show that ak 2: -bk for all k and therefore ak :::; bk as required. 4.
0
5.
First note that (ae + bd)(ad +be) 2: (ab- ed)2. To see this, we may assume a 2: e 2: d 2: b since a+b = e+d. Then ed- ab 2: Also we have the two obvious inequalities ae+bd 2: ed- ab and ad+be 2: ed-ab. Multiplying them together we get (ae + bd)(ad+be) 2: (ab- cd)2. Next (a2+e2)(a2+d2)(b2+e2)(b2+d2) 2 = ((ae+ bd)(ad+be)- (ab- ed)2) +(ab- ed)2 + ((a+b)2(e+d)2- 1) (ab- ed)2 2 = ((ae+ bd)(ad+be)- (ab- ed)2) +16(ab- ed)2
0.
<; =
cac+bd :ad+be)' - (oh- cd)2r +16(ab- cd)2 ca b)'�c+d)' - (ab- cd)2r+16(ab- cd)'
by AM-GM
+
(4 - (ab- ed)2)2+16(ab- ed)2. This final expression is an increasing function of (ab- cd)2. The largest value of (abed)2 is 1 when (a, b, e, d) = (1, 1, 2), (1, 1, 2, 2, 1, 1), (2, 1, 1). Consequently, 2 (4 - (ab- cd)2) +16(ab- ed)2:::; 25, proving the inequality. =
0,
0 ), (0,
39
0,
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) (Open Section, First Round)
Wednesday, 30 May 2012
2012
0930-1200 hrs
Instructions to contestants
1. Answer ALL 25 questions.
2. Write your answers in the answer sheet provided and shade the appropriate bubbles below
your answers.
3. No steps are needed to justify your answers.
4. Each question carries 1 mark. 5. No calculators are allowed.
PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO
40
Throughout this paper, let LxJ denote the greatest integer less than or equal to x. For example, ) 2 . 1 = 2 , 3.9 = 3 This notation is used in Questions 2 , 10, 16, 1 7, 18 and 22 .
L J
L J (
1 . The sum of the squares of 50 consecutive odd integers is 300850. Find the largest odd integer whose square is the last term of this sum. 1000 2. Find the value of
L Llog2 kJ . k=3
. 2 ) 3 . Given that f x is a polynomial of degree 201 2 , and that f k = k for k = 1 , 2 , 3, ·
(
x
find the value of 2014
()
( )
f 2014 .
(
· ·,
2013,
)
4. Find the total number of sets of positive integers x , y, z , where x , y and z are positive integers, with x < y < z such that x + y + z = 203. 5. There are a few integers n such that n 2 + n + 1 divides n 2 01 3 + 6 1 . Find the sum of the squares of these integers.
( ) 6. It is given that the sequence an �= l ' with a 1 = a2 = 2 , is given by the recurrence relation
for all n = 2 , 3, 4,
· · ·.
Find the integer that i s closest t o the value of
201 1 a + l k L2 a k . k=
7. Determine the largest even positive integer which cannot be expressed as the sum of two composite odd positive integers. 8 . The lengths of the sides of a triangle are successive terms of a geometric progression. Let A and C be the smallest and largest interior angles of the triangle respectively. If the shortest side has length 16 em and sin A - 2 sin B + 3 sin C sin G - 2 sin B + 3 sin A find the perimeter of the triangle in centimetres.
19 9'
9 . Find the least positive integral value of n for which the equation x +X +
y �
has integer solutions x 1 , x 2 , x 3 , · ·
(
···
+ X = 200 2 2 00 2
· , Xn ·
)
41
�
an be a real root of the cubic equation nx 3 + 2x - n
= 0, where n is a positive integer. 01 3 2 1 If f3n= l( n + 1)anJ for n= 2 3 4 · · · , find the value of f3k 1006 L · k=
10. Let
2
1 1 . In the diagram below, the point D lies inside the triangle such that and 90° . Given that 5 and 6, and the point M is the midpoint of find the value of 8 x DM 2 .
AC,
LBDC=
AB=
BC=
ABC
LEAD= LBCD
1 2 . Suppose the real numbers x and y satisfy the equations x 3 - 3x 2 + 5x
=1
Find x + y .
and y 3 - 3y 2 + 5y
=5
13. The product of two of the four roots of the quartic equation x4 - 18x 3 +kx 2 +200x- 1984 is -32. Determine the value of k.
=0
14. Determine the smallest integer n with n ?: 2 such that
is an integer. 1 5 . Given that f is a real-valued function on the set of all real numbers such that for any real numbers a and b, f af b ab
( )
Find the value of f 201 1 .
( ( ))=
16. The solutions to the equation x 3 - 4 x
l J= 5, wherek x is a real number, are denoted by XI X 2 X 3 . . . X k for some positive integer k. Find L x r . i=l
1 7. Determine the maximum integer solution of the equation
= 1001 + l !E_1 ! J + l !E_J l !E_3 ! J + . . . + l _!!_ 10! _ J 2! 42
18. Let A, B , C be the three angles of a triangle. Let L be the maximum value of sin 3A + sin 3B + sin 3C . Determine l10LJ. ( 19. Determine thex2number of sets of solutions x , y, z) , where x , y and z are integers, of the 2 2 2 2. + the equation
y +z
=
x y
20. We can find sets of 13 distinct positive integers that add up to 2142. Find the largest possible greatest common divisor of these 13 distinct positive integers. 21. Determine the maximum number of different sets consisting of three terms which2 form an arithmetic progressions that can be chosen from a sequence of real numbers a1 , a , · · · , a1 01 , where
22. Find the value of the series
� l 20121 + 2 k L...J
k=O
2 k+ 1
j.
23. The sequence (xn )�=1 is defined recursively by X n+ l with x1
=
=
(2 - J3) 1 (2 - J3) Xn + - Xn
'
1. Determine the value of X1001 - X401 ·
24. Determine the maximum value of the following expression
- ' ' -$ 1 - X 2 - X 3 - X4 - ' ' - X 2 0 1 4 -
where x1 , x 2 , · · · , x 2 014 are distinct numbers in the set
1 00 6 . -1 ). 25. Evaluate 22011 L ( -1) k 3k (2012 2k k=O
43
-1, 2, 3, 4, · · · , 2014---:
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2012 (Open Section, First Round Solutions)
1. Answer: 121 Let the integers be X + 2 X + 4
·· ·
100. Then ( X + 2) 2 + ( X + 4 )2 + + ( X + 100) 2 = 300850 Let y = X + 51 and regrouping the terms, we obtain [ (y - 49 )2 + (y + 49 )2 ] + [ (y - 47) 2 + (y + 47) 2 ] + . . . + [ (y - 1) 2 + (y + 1) 2 ] = 300850 which simplifies to 50y2 + 2 ( 1 2 + 32 + 52 + 72 + . . . + 49 2 ) = 300850 . the fact that 1 2 + 32 + 52 + · · · + (2n - 1) 2 = 34 3 - 31 n, we obtam. y = 72. Hence Usmg X = 21, so that the required number is 121.
Solution.
X+
···
n
D
2. Answer: 7986 Note that 2k+1 - 2k = 2k , and that 2k t 2k+ 1 if and only if llog2 tJ = k. Then the requires sum (denoted by S) can be obtained by 1 023 = 2: 2: llog2 tj + llog2 3j - L llog 2 tj k(=2 2k 9<2)k+l 10 {; k2k + 1 - �23 9 8192 + 1 - 23(9) = 7986 :::;
Solution.
<
9
s
t=lOOl
9
t
l
D
3. Answer: 4 Let g(x) = xf(x) - 2, hence g(x) is a polynomial of degree 2013. Since g(l) = g(2) = g(3) = = g(2013) = 0, we must have g(x) = >.(x - l)(x - 2)(x - 3) (x - 2012)(x - 2013) 2 Hence, for some >.. Also, g(O) = -2 = ->. 2013 ! , we thus have >. = 20131 g(2014) =20�3! (2013! ) = 2014 . f(2014) - 2 concluding that 2014 f(2014) = 4 Solution.
···
···
·
·
D
44
4.
Answer: 3333 . . mteger . sets = 20301 positive First note that there are (2022 ) = 202(201) 2 ( x, y, z) which satisfy the given equation. These solution sets include those where two of the three values are equal. If x = then 2x+z = 203. By enumerating, z = 1, 3, 5, , 201. There are thus 101 solutions of the form (x, x, z). Similarly, there are 101 solutions of the form (x, y, x) and ( x, y, y). Since x y z, the required answer is
Solution.
y,
·
<
·
·
<
D
5. Answer: 62 Since n 3 - 1 = (n7 - 1)(n2 + n + 1), we know that n2 + n + 1 divides n3 - 1. 2 0 13 Also, since n - 1 = (n3 ) 6 1 - 1, we also know that n2 + n + 1 divides n20 13 - 1. As 2 2 n 0 1 3 + 61 = n 0 1 3 1 + 62, we must have that n2 + n + 1 divides n20 13 + 61 if and only if n2 + n + 1 divides 62. Case (i): If n2 + n + 1 = 1, then n = 0, -1. Case (ii): If n2 + n + 1 = 2, there is no integer solution for n. Case (iii): If n2 + n + 1 = 31, then n = 6, -5. Case (iv): If n2 + n + 1 = 62, there is no integer solution for n. Thus, all the integer values of n are 0, -1, 6, -5. Hence the sum of squares is 1 + 36 + 25 = 62. Solution.
_
D
6. Answer: 3015 The recurrence relation can be written as l a:: - a:: l = ( n � 1 - �) - ( � - n : 1 ) Summing for n = 2 to N, we obtain Solution.
showing that
·
3 1- 1 . -;;:;,;- - 2 - ( N N + 1 ) Summing this up for N = 2 to N = 2011, we obtain 2 0 11 -'1-1 � (2010) - ( � - -1- ) = 3014. 5 + -1ak S = '"""' = � ak 2 2 2012 2012 showing that the integer closest to S is 3015. aN + l
_
45
D
7. Answer: 38 Let n be an even positive integer. Then each of the following expresses n as the sum of two odd integers: n = (n - 15) + 15 (n - 25) + 25 or (n - 35) + 35. Note that at least one of n - 15 n - 25 n - 35 is divisible by 3 so that it is composite, and hence n can be expressed as the sum of two composite odd numbers if n 38. Indeed, it can be verified that 38 cannot be expressed as the sum of two composite odd positive integers. Answer: 76 2 Let the lengths of the sides of the triangle in centimetres be 16, 16r and 16r 2 (where r 1). Then 1r2--2r2r++3r3 -199 so that r = -32 . Hence, the perimeter of the D triangle = 16 ( 1 + � + � ) = 76cm 9. Answer: 4 Since 2002 4(mod9) 43 1(mod9) and 2002 = 667 3 + 1, it follows that 20022002 4667x3+ 1 4(mod9). Observe that for positive integers x, the possible residues modulo 9 for x3 are 0 ±1 . Therefore, none of the following X� X� + � � + X� + � can have a residue of 4 modulo 9. However, since 2002 = 103 + 103 + 13 + 13, it follows that 20022002 2002 . (2002667 ) 3 (103 + 103 + 1 3 + 1 3 ) (2002667) 3 (10 . 2002667 ) 3 + (10 . 2002667 ) 3 + (2002667 ) 3 + (2002667 ) 3 This shows that x� + x� + x� + x! = 20022002 is indeed solvable. Hence the least integral value of n is 4. 10. Answer: 2015 Let f(x) = nx3 + 2x - n. It is easy to see that f is a strictly increasing function for n = 2 3 4 Further, 1 ( n : 1 ) = n ( n : 1 ) 3 + 2 ( n : 1 ) - n =(n ; 1)3 ( -n2 + n + 1 ) < O for all n 2. Also, f(1) = 2 Thus, the only real root of the equation nx3 + 2x-n = 0 for n 2 is located in the interval n(�1 1 ) . We must have n < an < 1 n < (n + 1 )an n + 1 -n+1 so that f3n = l(n + 1)anJ = n for all n 2. Consequently, -1- "201(33 k =-1- "2013 =-1- . 2012 (2 + 2013) = 2015 1006 k=2 1006 k=2 1006 2 Solution.
>
D
8.
Solution.
>
Solution.
=
=
x
=
=
X
X
X
D
·
2:
·
· .
2:
>
0.
==?
<
2:
�
�
k
D
46
1 1 . Answer: 22 Solution.
Extend CD to E such that CD = DE . ' ' ', ' '', ' ' ' ', ' ' ' ' ' ' ' ' ' ' \ ' \ ' ' \
,,
,
' ' ' ' \ '
�-------=
c
6
It is clear that D.CDB and D.EDB are congruent . Hence EB = CB = and L_BED L_BCD . Thus, L_BED = L_BCD = L_BAD implies that the points B , D , A are E are concyclic. Given that L_BDC = goo , hence L_EDB = goo . BDAE is a cyclic quadrilateral with EB as a diameter. Thus, L_EAB = goo . In the right-angled triangle EAB , we have
Since D and M are the midpoints of EC and AC respectively, DM = 8 x DM2 = 22 .
1 2 . Answer: 2 Solution.
2 From x 3 - 3
2 and from y 3 - 3 y
x + 5x
+ 5y
=
=
(
=
� AE = �· Thus, D
1 , we have
x - 1 ) 3 + 2 (x - 1 ) = -2,
5 , we have
) y-1 3
+ 2 (y - 1 ) = 2 . Thus ) ( ) ) ) ( 0 ( x - 1 3 + 2 (x - 1 + y - 1 3 + 2 y - 1 )2) )( ) ( )2 = (x + y - 2 ) ( (x - 1 - (x - 1 y - 1 + y - 1 2 + (x + y - 2 ) )2) )( ) ( )2 = (x + y - 2 ) ( 2 + ( x - 1 - (x - 1 y - 1 + y - 1 . For any real numbers x and y, we always have ? )( ) ( )2 0 y-1 (x - 1 - (x - 1 y - 1 and thus x + y - 2 = 0, implying that x + y = 2 . (
=
+
47
�
D
13. Answer: 86 Let a, b, c, d be the four roots of x 4 - 18x 3 + kx 2 + 200x - 1984 = 0 such that ab = -32. Then a + b + c + d = 18, ab + ac + ad+ be + bd + cd = k, abc + abd + acd + bed = -200, abed = -1984. Since ab = -32 and abed = -1984, we have ed = 62. Then, from abc + abd + acd + bed = -200 we have -200 = -32c - 32d + 62a 62b = -32(e + d) + 62(a + b). Solving this equation together with the equation a + b + c + d = 18 gives that a + b = 4, c + d = 14. From ab + ac + ad + be + bd + ed = k, we have
{
Solution.
+
= 30 + (a + b)(c + d) = 86.
D
14. Answer: 337 Solution.
Assume that
and so
+
(n 1)(2n + 1) = 6m2 . Thus 6 (n + 1)(2n + 1), implying that n 1 or 5 ( mod 6). = 6k + 5 . Then m 2 = (k + 1)(12k + 11). Since (k + 1) and (12k + 11) are relatively prime, both must be squares . So there are positive integers a and b such that k + 1 = a2 and 12k + 11 = b2 . Thus 12a 2 = b2 1. But, as 12a 2 O ( mod 4) and b2 + 1 1 or 2 (mod 4), there are no integers a and b such that 12a2 = b2 + 1. Hence Case 1 cannot happen. n = 6k + 1. Then m 2 = (3k + 1)(4k + 1). Since 3k + 1 and 4k 1 are relatively prime, both must be squares. So there are positive integers a and b such that 3k + 1 = a 2 and 4k + 1 = b2 . Then 3b2 = (2a - 1)(2a + 1). Observe that in the left-hand side, every prime factor except 3 has an even power. So neither 2a - 1 nor 2a + 1 can be a prime other than 3. Now we consider positive integers a such that neither 2a - 1 nor 2a + 1 can b e a prime other than 3. If a = 1, then b = 1 and n = 1. So we consider a 2. The next smallest suitable value for a is 13. When a = 13, we have 3b2 = 25 27 and so b = 15, implying that k = 56 and so = 6k + 1 = 337. =
Case 1: n
+
=
=
Case 2 :
+
2:::
X
n
48
D
15. Answer: 2011
f(f(1)f(b))
f(b)
f(b)
From the recurrence relation, = f( 1 ) b and f(f(b)f( 1 ) ) = 1. Hence, = f( 1)b. By letting = / ( 1 ) , we obtain f(f( 1 ) ) = (f( 1 ) ) 2 . Also, from the given functional equation, we have f(f( 1 ) ) = 1, hence (! ( 1 ) ) 2 = 1 , following that / ( 1 ) is either 1 or - 1 . D Hence f(201 1 ) = 201 1 .
b
·
16. Answer: 19 <
�
2::
x - 1 LxJ x . Note that if x 3, there will be no solution as x3 - 4LxJ x3 - 4x = x(x2 - 4) 3(5) = 15. Also, if x - 2 , there will be no solution as x 3 -4 L xJ x3 -4(x - 1) = x(x 2 - 4) + 4 4 . Hence the solution must be in the interval ( - 2 , 3) . If LxJ = - 2 , then x3 = -3, giving x = N, which is a solution. If LxJ = - 1 , then x3 = 1, giving x = 1 which contradicts with LxJ = - 1 . LxJ = 0, then x3 = 5, hence there is no solution. If LxJ = 1 , then x3 = 9 . Since 2 W 3, there is no solution. If LxJ = 2, then x3 = 13. Since 2 m 3, X = m is a solution. D Thus, the required answer is -3 + 13 = 10. 17. Answer: 584 Solution.
Note that
2::
2::
�
<
If
<
<
<
Solution.
�
<
It is clear that
l � J + l;,J + l:,J + . . . + l 1�! J
x,
x
is a monotone increasing function of and when = 6 ! , the above expression has a value larger than 100 1 . Thus each solution of the equation
l �,J + l;,J + l:,J + . . . + l 1� ! J = 100 1
x is a solution, then l �,J + l;,J + l:,J + . . . + l 1� ! J = l �,J + l;,J + l :, J + l�J + l:,J .
is less than 6! and so if
As
x
<
6 ! , it has a unique expression of the form X
+ b 4! +
C X
X
C X
X
X
e,
+ d 2! + 1 . Note that where b, c, d, are non-negative integer with a 5, b 4, c 3, d 2 , if 3! + d 2 ! + X = a 5! + b 4! + then l�, J + l;,J + l:,J + l :,J + l � J = 206a + 41b + 10c + 3d + Since 41b + 10c + 3d+ 201 , we have 800 206a 1001 and so a = 4 . Thus 41b + lOc + 3d + = 1 77, X=a
a,
e
X
5!
3!
�
�
X
�
�
e
�
e,
e.
e
�
�
�
e
49
which implies that
c
b = 4 and so on, giving that = d = 1 and X
=4
X
5! + 4
X
4! + 1
X
3! + 1
X
e
= 0. Thus
2 ! + 0 = 584.
As 584 is the only integer solution, the answer is 584. D 18. Answer: 25 Solution.
We shall show that -2 ::::; sin 3A + sin 3B + sin 3C ::::; 3v'3/2.
Assume that A ::::; B ::::; C. Then A ::::; 60° . Thus sin 3A 2:: 0 . It is clear that sin 3B 2:: - 1 and sin 3C 2:: - 1 . Thus sin 3A + sin 3B + sin 3C 2:: -2. Let B = C. Then B = C = goo - A/2. If A is very small, B and C are close to goo , and thus sin 3A + sin 3B + sin 3C is close to -2. Now we show that sin 3A + sin 3B + sin 3C ::::; 3v'3/2. First the upper bound can be reached when A = B = 20° and C = 140° . Let X = 3A,
Y
= 3B and Z = 3(C - 120° ) . Then X + Y + Z = 180° and sin 3A + sin 3B + sin 3C = sin X + sin Y + sin Z.
Suppose that X, Y, Z satisfy the condition that X + Y + Z = 1 80° such that sin X + sin Y + sin Z has the maximum value. We can then show that X = Y = Z. Assume that
X ::::; Y ::::;
Z. If X < Z, then
X-Z . . . X+Z . X+Z sm X + sm Z = 2 sm cos < 2 sm , 2 2 2 implying that . . X+Z . X+Z . . y+� . y+� z < � +� �X + � 2 2
which contradicts the assumption that sin X + sin Y + sin Z has the maximum value. Hence X = Y = Z = 60° , implying that A = 20° , B = 20° and C = 140° and sin 3A + sin 3B + sin 3C = 3v'3/2. Since
y'3
�
1 . 732 , the answer is then obtained.
D
1 g . Answer: 1 Solution. Note that x = 0, y = 0 and z = 0 is a solution of this equation. We shall show that this is its only integer solution by proving that if x, y, z is a solution of this equation and whenever x, y, z are divisible by 2k , they are also divisible by 2k + 1 for any k 2:: 0.
Let
x
=
2k x' , y
= 2ky' and
z
= 2k z ' . Then
x2
+ y2 + z2 =
x2 y 2
is changed to
It is easy to verify that only when x' , y' , z ' are all even, x'2 + y '2 + z 12 and the same remainder when divided by 4. Thus x, y, z are divisible by 2 k + 1 .
50
2 2 k x12 y'2
have D
20. Answer: 21
Let d be the greatest common divisor ( gcd ) o f these 13 distinct positive integers. Then these integers can be represented as da1 , da2 , · · · , da13 , where gcd ( a1 , a2 , · · · , a13 ) = Let S denote a1 + a2 + · · · + a13 · Then Sd = In order for d to be the largest pos sible, S must be the smallest . Since S ;::: + + 3 + · · · + 3 = and that S divides and that = x 3 x 7 x the smallest possible value of S can be x = and the largest value of d is thus By choosing ( a1 , a2 , a3 , · · · , a12 , a13 ) = 2, 3, · · · , D we conclude that d = is possible. Solution.
13
1.
2142 2
1 2
51, 21.
21
2142.
1 91
2142, 2 51 102, (1, 12, 24),
21. Answer: 2500 Solution.
First , for the following particular sequence, there are really three-term arithmetic progressions which can be chosen from this sequence: They are s , and t with
s
s,
1, 2, 3, . . . ' 101. i with 1 i 51 and 2i - i, �
s
<
�
2500 different
i, 2i - for all integers for all integers i 52 i 101. Now we show that for any given sequence of real numbers a1 a2 aw1 , there are at most 2500 different three-term arithmetic progressions which can be chosen from this sequence. Let as , ai , at represent a three-term arithmetic progression. It is clear that 2 i 100. If i 51, then the first term as has at most i - 1 choices, as must be an index in 1, 2, i - 1 . If 52 i 100, then the third term at has at most 101 - i choices, as must be an index in i + 1, i + 2, , 101 . 2 �
�
< t �
�
···,
t,
<
�
< ··· <
�
s
�
t
�
t
· ··
So the number of different three-term arithmetic progressions which can be chosen from this sequence is at most 1 00 51 + = + 2 + . . . + 5o + + 2 + . . . + = 25oo . 2 2 i=5 i= D
I)i - 1) L (101 - i) 1
22. Answer: 20121 Solution.
Write
x
:= x
x
- L J . Then
1
49
if !!C2 < 12 otherwise
Thus, we have
Applying the above result for
x=
;,
f ( l ; J - l 2::1 J ) k=O l;J
In particular, when
n=
n.
20121, the infinite series converges to 20121. 51
D
23 . Answer: 0 Solution.
Let X n = tan O: n . Since 2 -
X n+I = tan a n + I =
-J3 = tan {;( ) , it follows that
(
tan an + tan ( � ) 1r = tan O:n + 1f 12 1 - tan an tan ( I 2 )
)
( So, X n+ l 2 = tan an + 1r ) = tan an X n , implying that this sequence has a period of 1 2 . Observe that 1 0 0 1 = 5 (mod 12) and 401 = 5(mod 12) . Consequently, =
Xl QO I - X40 1 = X5 - X5 = 0 D 24. Answer: 2013 Solution.
be 2014, as
First it is clear that the answer is an integer between 0 and 2014. But it cannot X I - X 2 - X 3 - X4 X 20 I4 and ·
·
·
·
X I + X2 +
·
·
·
·
·
-
+ X 2 0 I4 = 1
2+
+
·
·
·
+ 2014 = 1007
X
2015
have the same parity. Now we j ust need to show that 2013 can be achieved . For any integer k , ( 4k + 2) - ( 4k + 4) - ( 4k + 5) - ( 4k + 3) = 0 Thus 2 - 4 - 5 - 3 - . . . - ( 4k + 2) - ( 4k + 4) - ( 4k + 5) - ( 4k + 3) -2010 - 2012 - 2013 - 201 1 - 2014 - 1 =
0 - 2014 - 1
=
2013 D
25. Answer: 1 Solution. Consider the complex number binomial theorem one has
(
1f
1f
w
= cos
J + i sin J . On one hand, using the
) 20 I 2
Re cos 3 + i sin 3 20 I2 Re + i -;
( � V3)
(�)
c ) (�) ( ��::: ) [ ( ) ( )
2 0I 2
_
20 I O (; ) 2
o12 2
+...+
2012 1 1-3 2 2 I 0 2 2
_
20 12 + 4
( � ) ( �: ) 2 00 8
( )]
3 2 2012 + . . . + 3I 006 2012 + 201 2 4
52
On the other hand, using the De Moivre's theorem one has
( 20121f
Re ( w2 0 1 2 ) = Re cos Thus ,
+
i sin
20121f ) 3
_1_[1 - (2012) (2012) . . . 4 2 2012 (20124 ) . . . � [1 - ( 2 )
2 20 1 2 so that
3
2 20 11
3
+
3
32
+
32
- 3 1 004
+
+
53
= cos
20121f
(2012 2010)
- 3 1 00 4
3
+
2012) (2010
= cos
3 1 006
+
21f 3
=
2
-�
(2012 2012) ] - �2 2012) ] 1 (2012
3 1 006
=
=
D
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO ) 2012
(Open Section, Round 2) Saturday, 30 June 2012
1. The incircle with centre
0900-1300
of the triangle ABC touches the sides BC, CA and AB at and respectively. The line intersects the segment at K. Prove that A, K and M are collinear where M is the midpoint of BC.
D, E
F
2. Find all functions
f
:
IR
I
---+
ID
IR
EF
so that
+
+
=
(x y )(f(x ) -f (y )) (x-y ) f (x y ) for all
x, y
3. For each
i
E JR. =
1, 2, . . . , N, let ai, bi, ci be integers such that at least one of them is odd. Show that one can find integers z such that xai + ybi + zci is odd for at least 4N different values of i .
x, y,
/7
4.
Let p be an odd prime. Prove that ! P-2 + 2p -2 + 3p -2 + . . . +
5.
(� - 1 ) p -2
-
2 - 2P p
( mod p ) .
There are 2012 distinct points in the plane each of which is to be coloured using one of n colours so that the number of points of each colour are distinct. A set of n points is said to be multi-coloured if their colours are distinct. Determine n that maximizes the number of multi-coloured sets.
54
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO)
2012
(Open Section, Round 2 solutions)
1. A
c
B
Let the line AK intersect EC at M. We shall prove that M is the midpoint of EC. Since LFIK = LE and LEIK = LC, we have FK EK Also
FK sin LF AK
Therefore,
sin LFIK sin LEIK
AE sin LAKE
AF sin LAKF sin LFAK sin LKAE
sin E sin C '
FK EK
EK . sin LKAE
sin E sin C '
Consequently, EM EM . AM sin LFAK sin C = = = CM AM CM sin E sin LKAE so that EM = CM.
55
l,
2. Suppose that
f is a solution. Let a = 21 (! (1) - f (-1)), b = 21 (! (1) +! (-1)) and g (x ) = f (x ) - ax - bx2. Then (x +y )(g(x ) - g (y )) = (x - y)g (x +y ) and g (1) = g( -1) = 0. Letting y = 1 and y = -1 above give (x + 1) g (x ) = (x - 1) g (x + 1) xg (x +1) = (x +2)g(x) .
Thus
x(x +1)g (x) = x(x - 1) g(x +1) = (x - 1)(x + 2) g (x) for all x. So g (x ) = 0 for all x. Hence f(x) = ax+bx2. We can check directly that any function of this form (for some a, b E JR.) satisfies the given equation. Consider all the 7 triples where z are either 0 or but not all 0. For is odd. Thus among the 7 sums each at least one of the numbers 3 are even and 4 are odd. Hence there are altogether 4N odd sums. Thus there is choice of for which at least 4N 7 of the corresponding sums are odd. You can think of a table where the rows are numbered . . . , N and the columns correspond to the 7 choices of the triples The 7 entries in row are the 7 sums Thus there are 4 odd numbers in each row, making a total of 4N odd sums in the table. Since there are 7 columns, one of the columns must contain at least 4N 7 odd sums.
(x, y, z), x, y, ai, bi, Ci / 1, 2, i (x, y, z) .
3.
i, (x, y, z)
1
( xai +ybi +zci. / )
i 1, 2, . . P, ;1 , 2i ( p ) = (p -1) (p - 2) . . . (p - (2i -1)) (-1) (-2) (- (2i - 1)) p 2i (2i - 1) ! (2i - 1) !
4.
·
First, for each =
·
Hence
(p-1)/2
L i= 1
xai+Ybi+zci,
·
·
_
_
1 (mod p) .
. ( ) 2 (p-1)/2 ·p-1 ( ) 2 - L 2i. p 2i = p i=1 i= 1 (p- 1)/2 ( p ) 2 --p (mod p) (by Fermat 's Little Theorem. ) = i 1 2i
·p-2 = 't
(p-1)/2
_
-
·p- 2 't
't
P
_
-
-
� L....,;
't
P
� L....,;
=
The last summation counts the even-sized nonempty subsets of a p-element set, of which there are P - 1
2 - 1.
56
5.
Let m1 < m2 < < mn be the number of points of each colour. We call m1, m2, . . . , mn the colour distribution. Then m1 + + mn = and the num ber of multi-coloured sets is M = m1 m2 . . . mn. We have the following observations. ·
·
·
·
·
2012
·
(i) m1 > 1 . For if m1 = then m1m2 . . . mn < m2m3 . . . mn - 1(1 + mn ) This means if we use n- colours with colour distribution m2, m3, . . . , mn_ 1, + mn), we obtain a larger M.
1,
1
·
(1 (ii) mi+1- mi ::::; 2 for all i . For if there exists k with mk+1 - mk 2': 3, then the colour distribution with mk, mk+1 replaced by mk 1, mk+1- 1 yields a larger M. (iii) mi+1- mi 2 for at most one For if there exist i j with mi +1 - mi mj+1 - mj 2, the colour distribution with mi, mj+1 replaced by mi 1, mj+1 - 1 yields a larger M. (iv) mi+1 - mi 2 for exactly one For if mi +1 - mi 1 for all then m1 mn nm1 n(n2-1) 2012 4 503. Thus n(2m1 - 1 n) 8 503. Since 503 is prime, the parity of n and 2m1 -1 n are opposite and 2m1 - 1 n n, we have n 8 and m1 248. The colour distribution with m1 replaced by two numbers 2, 246 (using n 1 colours) yields a larger M. 2. If mn - mn 1 2, then from (iv) , we have m1 mn (v) m1 n(n-1) 1 2012. Thus -n(2m1 -1 ) 2 2011. Since 2011 is prime, we get nm1+ 2 n 2 and m1 1005 which will lead to a contradiction as in (iv) . Thus mn-mn_ 1 1. mi+1- mi 2 for some 1 ::::; i ::::; n - 2. Suppose m1 2': 3. Let m' mi+2 - 2. Then mi m' mi+1 with replacing mi+2 by 2, m' yields a larger M. Thus m1 2. From the above analysis, with n colours, we see that the colour distribution 2, 3, 2, . . ., n 1, n 2, with 3 ::::; i ::::;2 n, yields the maximum M. Now . . . , i- 1, i 1, we have 2: mi � (n 1) ( n 4) - i 2012. Thus n + 5n- 4020 2i, 3 :=::; i ::::; n, i.e. , n2 5n 2': 4026 and n2 3n ::::; 4020. Thus n 61 and i 3. Thus the maximum is achieved when n 61 with the colour distribution 2, 4, 5, 6, . . . , 63. =
·
·
·
+
=
+
=
=
<
+
<
=
+
+
i.
=
i.
=
=
=
=
=
+
+
=
=
<
·
=
+
=
=
+
+
n
=
+
=
i,
+
=
+
+
+
+
=
57
+
>
·
·
·
+
=
·
+
=
·
+
=
i +
=
=
=
=
=
Multiple Choice Questions 1. Calculate the following sum: 1 2 3 4 10 -+ -+ -+ -+ . . . + . 16 210 2 4 8
(A)
50 3 . 256'
507
505
509
(B) 5 ,. C ; (D) 5 ,. (E) None of the above. 2 6 ( ) 256 2 6
2. It is known that the roots of the equation x5 + 3x4- 40441 18x 3 - 12132362x 2- 12132363x- 201 12= 0
are all integers. How many distinct roots does the equation have? (A) 1;
(B) 2; ( C) 3; (D) 4; (E) 5.
3. A fair dice is thrown three times. The results of the first, second and third throw are recorded as x, x,
y
(A)
y
and
and 1 ; 12
z
z,
respectively. Suppose x + y=
z.
What is the probability that at least one of
is 2? 8
(B) � , ( C ) 5 ; (D) 1 8 ·
1 3;
(E)
!__. 13
4. Let
= 1 000 · · · 000 1 000 · · · 000 50 . '-....--" '-....--" 2012 times 2011 times
X
Which of the following is a perfect square? (A) x- 75;
(B) x- 25; (C) x ; (D) x + 25; (E) x + 75.
5. Suppose N 1, N2, ..., N2o11 are positive integers. Let
X= (N1 + N2 + · · · + N2o10) (N2 + N3 + · · · + N2ou ), Y= (N1 + N2 + · · · + N2ou) (N2 + N3 + · · · + N2o1o) . Which one of the following relationships always holds? (A) X=Y;
(B) X>Y; ( C ) X
. 6. Consider the following egg shaped curve. A BC D is a circle of radius 1 centred at 0. The arc AE is centred at C , CF is centred at A and EF is centred at D. 2
(A) 1680;
(B) 1712; (C) 3696; (D) 3760; (E) None of the above.
10. In the set {1, 6, 7, 9}, which of the numbers appear as the last digit of nn for infinitely many
positive integers n? (A) 1, 6, 7 only;
(B) 1, 6, 9 only; (C) 1, 7, 9 only; (D) 6, 7, 9 only; (E) 1, 6, 7, 9.
Short Questions z x y 11. Suppose -+ -+-= a b c
12. Suppose x=
a b c . ..J2 and-+-+-= 0. Fmd
13 v'19 + sv'3
x
y
z
. Find the exact value of x4- 6x 3- 2x 2 +18x + 23 x 2- 8x + 15
13. Let a1= 3, and define an+l=
1 a J3 n � for all positive integers n. Find a2011 · an + 3
14. Let a, b, c be positive real numbers such that a2 + ab + b 2= 25, b2 + be+ c2= 49, c2 + ca + a2= 64.
4
distinct ways of using two tiles of size 1
x
1, two tiles of size 1
x
2 and one tile of size 1
x
4.
It is not necessary to use all the three kinds of tiles. )
24. A 4 2
x
x
4 Sudoku grid is filled with digits so that each column, each row, and each of the four
2 sub-grids that composes the grid contains all of the digits from 1 to 4. For example,
Find the total number of possible 4
x
4
3
1
2
2
1
3
4
1
2
4
3
3
4
2
1
4 Sudoku grids.
25. If the 13th of any particular month falls on a Friday, we call it
Friday the 13th.
It is known
that Friday the 13th occurs at least once every calendar year. If the longest interval between two consecutive occurrences of Friday the 13th is
x
months, find
x.
26. How many ways are there to put 7 identical apples into 4 identical packages so that each package has at least one apple?
27. At a fun fair, coupons can be used to purchase food. Each coupon is worth $5, $8 or $12. For example, for a $15 purchase you can use three coupons of $5, or use one coupon of $5 and one coupon of $8 and pay $2 by cash. Suppose the prices in the fun fair are all whole dollars. What is the largest amount that you cannot purchase using only coupons?
28. Find the length of the spirangle in the following diagram, where the gap between adjacent parallel lines is 1 unit.
6
E
D
c
34. Consider an equilateral triangle A BC , where A B = BC = CA = 2011. Let P be a point
inside LA BC . Draw line segments passing through P such that DE
II BC , FG II CA and
HI II AB. Suppose DE : FG : HI= 8 : 7 : 10. Find DE +FG + HI . A
c
35. In the following diagram, A B
j_
BC . D and E are points on segments A B and BC respec
tively, such that BA + AE = E D + DC . It is known that A D = 2, BE = 3 and EC = 4. Find BA + AE .
8
Note that X= (10 1012 + 1) X 10 2014 +50= 104026 + 102014 +50 = (102013 )2 +2 X 102013 X 5 +50= (102013 +5) 2 +25. So x- 25 is a perfect square. 5. Answer: ( B ) .
Let K= N2 + · · · +N2010· Then X= (N1 +K) (K +N2on) andY= (N1 +K +N2on)K.
6. Answer: ( A ) .
7r (22 ) ,---... The area enclosed by A D , DE and AE is 8 -
The area of the wedge E DF So the area of the egg is:
. 7r (2- J2)2 = IS 4
(
-
1f - 1. 1= 2
)
3 J2 2-
1f
.
� + 1 +2 (�- 1) + (�- J2) 1r= (3- J2)1r- 1. x
7. Answer: ( B ) .
The left shows that 3 colours are not enough. The right is a painting using 4 colours.
8
.
Answer: (E). Since 5 I (24- 1), 71 (36- 1), 11 I (510- 1), 13 I (7 12- 1),
9. Answer:
n
is divisible by 5, 7, 11 and 13.
(C).
We consider the position of the Black Knight. The number of positions being attacked by the White Knight can be counted.
10
13. Answer: 3.
J33 x 01 . Then f ( (x))= Let f (x)= J X+ 3
vfsx-1 _J3 = J (J (J (x)))= x+vfs J3 vfsx 1 + 1 x+vfs
Since 2010= 6
x
J3 vs-1 _1
�vis 3x-1 +J3 x+vfs
3x-J3 - x-J3 J3x - 1 + V3x+ 3
-
x-J3 J3x+ (
3x - 1- v3x- 3 _.!. v/"il . So J (J (J (J (J (J (x))))))= x. = X 3x-J3 + x/"il+J3
335, a2o11= f (J (J··· f (J (3)) ·· ))= 3. '-----v--" 2010 copies ·
14. Answer: 129. Consider the following picture, where LAOB = L BOC= LCOA= 120°, OA= a, OB= b and OC= c. A
c
B
Then IBCI= 5, I CAl= 7 and lA B I= 8. The area of the triangle A BC is yl10 (10- 5) (10 - 7) (10 - 8)= 10.J3. 1V3 (ab +be+ ca)= 10J3. So ab +be+ ca= 40. Then 2 2 2 (a + b + c) 2= (a2+ ab + b2) + (b2 +be+ c2) + (c2 + ca + a2) + 3 (ab +be+ ca)= 258. Thus, (a+ b + c) 2= 129. 15. Answer: 0. Define Q (x) = (1 + x)P (x) - x. Then Q (x) is a polynomial of degree 2011. Since Q(O) = Q (1)= Q (2)= ···= Q (2010) = 0, we can write, for some constant A, Q (x)= Ax(x- 1) (x- 2)··· (x- 2010). 1= Q (-1)= A (-1) (-2) (-3)·· · (-2011)=-A· 2011!. Then Q (2012) = A· 2012!= -2012, Q (2012) + 2012 and P ( 2012)= = o. 2013 16. Answer: 9241. 12
Suppose x 1
=
x2
M
=
=
2:: >
2::
2::
=
=
Therefore, M
�
···
=
=
Xk
1
=
�
2
<
Xk+l
�
···
�
X2011 · Let M
=
x 1 · · · X20l1 · Then
Xk+1Xk+2 . . . X2QlOX20ll (Xk+l- 1)x k+2 · · · X2010X2011 +Xk+2 . . . X2010X2011 (Xk+l- 1) 2 +Xk+2 · · · X2010X2011 .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
(x k+l- 1) 2 + · · · + (x 2009- 1)2 +X2010X2011 (x k+l- 1) 2 + · · · + (x 2009- 1)2 + (x 2o10- 1)2 + (x 2o11- 1) 2 2 (x k+l + · · · +x 2o11- (2011- k)) 2 (M- 2011) .
4022. On the other hand, (1, 1, . . . , 1, 2, 2011) is a solution to the equation.
So the maximum value is 4022. 2 1. Answer: 101. 2::
If n
M (91)
102, then M (n) =
M (M (102))
For each k
=
=
=
n- 10
M (92)
2::
92.
M (M ( 103))
=
1, . . . , 10, M (80 + k)
=
M (70 + k) M(k)
=
=
=
M (93)
M (M (91 + k))
=
M (M(81 + k))
···
M (91)
=
M (M(11 + k))
=
=
Hence, all integers from 1 to 101 are solutions to M (n)
=
=
M (91) M (91) =
=
=
M ( 101)
=
9 1.
9 1, and thus 9 1, 9 1.
91.
22. Answer: 19.
20 + 1 1 . (M) n 1 2on+l + 1 1n+l · n n - (n +1) (1 +(M) n) n +1 20 +1 1
An+l An
Then An+l
<
An if n
Note that 10 + If 10 n
�
�
n
�
>
�
10 +
11 ) n 1 + 20
18, then n
18 implies An
<
�
<
n ; and An+l 1 + 11 w)
10 +9
10 +8
An+l ·
�
<
=
19. Son
10 +
2::
An if n
>
<
19 implies An
9 . ; 1f n n 11 1 + ( 20 )
<
10 + >
1+
�!! w
)n
.
An+l ·
10, then n
<
10 +
9 n Hence, 1 + ( 11 20 ) .
23. Answer: 169.
Let an be the number of ways to pave a block of 14
1
X
n. Then an
=
an-1 +an-2 + an-4 with
Case 2:
Case 3:
The first year is a leap year. Jan
Feb
Mar
Apr
Mar
Jun
Jul
Aug
Sep
Oct
Nov
Dec
0
3
4
0
2
5
3
6
1
4
6
2
5
5
1
3
6
@] 1
4
@]
2
5
0
The second year is a leap year. Jan
Feb
Mar
Apr
Mar
Jun
Jul
Aug
Sep
Oct
Nov
Dec
0
3
3
6
1
4
2
5
5
4
5
1
3
[]]
4
@
@
3
1
[]]
5
0
1
2
From these tables we see that the answer is 14. The longest time period occurs when the Friday of 13th falls in July of the first year and in September of the second year, while the second year is not a leap year. 26. Answer: 350. By considering the numbers of apples in the packages, there are 3 cases: l) (4, l, l, l). 2) (3, 2, 1, 1 ) . 3) (2, 2, 2, l).
X 6 X 5 X 4 = 35. 4x3x2xl
(7) 7 (7) ( ) 7 � (7) ( ) ( ) 4
=
3
3! 2
4 2
=
5 2
6X5 = 3x2xl X
3 2
=
�
6
X
X
7 2
4 X3 = 35 2xl
X
6 = 210.
4 1
X
3 2
X X
6 1
X
5 2
X X
2 = l05. 1
X X
So the total number of ways is 35+ 210 + 105 = 350. 27. Answer: 19. Note that 8 + 12 = 20, 5 + 8+ 8 = 21, 5 + 5 + 12 = 22, 5 + 5 + 5 + 8 = 23, 8+ 8 + 8 = 24. If 2::
n
2::
25, write
n=
5k + m where 20
:::;
m
:::;
24 and k is a positive integer. So any amount
25 can be paid exactly using coupons.
However, 19 cannot be paid exactly using these three types of coupons. 28. Answer: 10301. . The broken line is constructed using
"
L ", with lengths 2, 4, 6, . . . , 200.
100 + 101 = 201. Then the total length is 2 (1 + 2 + 3 + 29. Answer: 15. 16
·
·
·
The last
+ 100) + 201 = 10301.
"
L " is
So c= 12 or c=-20 (ignored ) . Substitute y=
-x
+ 12 into the parabola:
2=
x
-x
+ 12
=*
= 3,-4. So A is (3, 9) and B is (-4, 16). Then
x
I A BI 2= (3- (-4)) 2+ (9- 16) 2= 98. 33. Answer: 30. Draw BF
..l
LE= 30°.
J2 CE , where F is on CE . If A B = 1, then BF = 2 and BE =
J2. Thus
E
D
34. Answer: 4022. Set D P= GP= a, I P= FP= b, E P= HP= c. Then DE +FG+HI= (a+ c)+ (a+ b) + (b + c)= 2 (a + b+ c)= 2 35. Answer: 10. By given, B D + 2 +AE= B D + DC. So 2 +AE= DC . Note that A B 2+ BE 2= AE 2 and B D 2+ BC 2 (2 + B D ) 2+ 32= AE 2, 4 (AE + B D )= 32. Then AE + BA
=
=
DC 2. Then
B D 2+ 72= (AE + 2) 2.
32 4 + 2= 10.
18
x
2011= 4022.
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO)
2011
(Junior Section, Round 2 solutions)
1. It's equivalent to prove x2y2 :;:::: (ac + bd)2 as all the numbers are nonnegative. This
is true s ince
x2 y 2
(a2 + b2)(c2 + d2) (ac)2 + (bd)2 + a2d2 + b2c2 :;:::: (ac)2 + (bd)2 + 2(ac)(bd) AM-GM (ac + bd)2.
= =
=
Let the tangent at P meet the tangent at R at the point S. Let 0 be the centre of r1. Then OR ST is a s quare. Hence LKPR L R P S 45°. Als o LNPS LNKP LPMS LMPK. Thus LMPR LRPN. 2.
=
=
=
=
=
=
K
3. Let ai 0,
maxSi, bi minSi and s uppos e that t1 then a1 :;:::: b j. Therefore a1 E Sj. =
=
=
min{ti}. For each j, if 81 n Sj i=
Note: Problem 4 in the Senior Section is the general vers ion. 4.
Replace 2011 by any pos itive odd integer n. We firs t s how by induction that am 3m2n-m - 1 for m 0. Suppos e 0, 1, ..., n- 1. This is certainly true for m + + nit's true for s ome m < n - 1. Then 3am 1 3m l2 m - 2. Since n- m > 1, the odd part is 3m+l2n-m-l - 1 which is am+l· Now an-1 3n-121 - 1. Thus =
=
=
=
=
20
Singapore Mathematical Society
Singapore Mathematical Olympiad (SMO)
2011
(Senior Section)
Tuesday,
31
May
2011
0930-1200
hrs
Instructions to contestants 1. Answer ALL 35 questions. 2. Enter your answers on the answer sheet provided. 3. For the multiple choice questions, enter your answer on the answer
sheet by shading the bubble containing the letter (A, B, C, D or E) corresponding to the correct answer.
4. For the other short questions, write your answer on the answer sheet and shade the appropriate bubble below your answer. 5. No steps are needed to justify your answers. 6. Each question carries 1 mark. 7. No calculators are allowed.
PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO
22
6. Determine the value of
2
1 1 + 0+Vfs+2 0-V'8+2 (A) 4- 0 (E) 40-2 7. Let
x
=
is true?
(A)
(D)
1
(B) 2-20
1
log� 2
+
1
1
logi 4
1. 5 < X < 2
3 <X< 3. 5
+
(C) 4 +0
1
1.
log� 8
(D)
20 + 4
Which of the following statements
(B) 2 < x < 2. 5 (E) 3.5 <X<4
(C) 2.5
<X<
3
8. Determine the last two digits of 756•
(A) 01
(B) 0 7
9. It is given that
x
(C) 09
(D)
43
(E) 49
and y are two real numbers such that
y > 1. Find the smallest possible value of
x
>
1
and
=
c-1
logx2011 +logy2011 logxy 2011
(A) 4 10.
(B)
6
(C)
8
(D)
(E) 12
10
It is given that a, band care three real numbers such that a+ b
and ab
(A)
5
=
c2 - 7c +14. Find the largest possible value of a2 + b2.
(B)
6
(C)
8
(D)
9
(E) 10
Short Questions 11.
Find the value of
20112 +21112-2 X 2011 X 2111 25 24
18.
In the diagram below, the lengths of the 2three2 sides of the triangle are a +2 b = 2011. Find the value a em, b em and c em . It is given that of
c
cot C
cot A +cotE
b c a
19.
Su p pose there are a total of
petition, and at least
1011
partici pants, at
2011
partici pants in a mathematics com
1000 of them are female . Moreover, given any least 11 of them are male . How many male
partici pants are there in this com petition?
20.
Letf : Q \
{0, 1}
-t
Q be a function such that
2
xf ( x)+f for all rational numbers
2 (X -1) = 2x x
x #- 0, 1.
Here Q denotes the set of rational
�
numbers . Find the value off( ).
21.
In the diagram below , ABCD is a convex quadrilateral such that AC
intersects BD at the midpoint E of BD. The point His the foot of
the per pendicular from A onto DE, and Hlies in the interior of the segment DE. AH=
40
Su p pose LBCA =
em and CD=
x
90°, CE = 12
em . Find the value of A
c
26
em, EH=
x.
15
em,
28. It is given that
a,
b,
c
and d are four positive prime numbers such
that the product of these four prime numbers is e qual to the sum of
55 consecutive positive integers. Find the smallest possible value of
a+ b+ c +d. ( Remark: distinct. )
The four numbers
a,
b,
c,
d are not necessarily
29. In the diagram below, ABC is a triangle with AB= 39 em , BC = 45 em and CA= 42 em. The tangents at A and B to the circumcircle of
DABC meet at the point P. The point D lies on BC such that PD
is parallel to AC . It is given that the area of DABD is x cm 2. Find the value of x.
30. It is given that
a and
b are positive integers such that
a has
exactly
9 positive divisors and b has exactly 10 positive divisors. If the least
common multi ple
( LCM ) of
a and
b is 4400, find the value of
a
+ b.
31. In the diagram below, ABCD is a quadrilateral such that .!_ABC= 135° and L_BCD= 120°. Moreover, it is given that AB = 2 J3 em,
BC= 4 - 2 -/2 em, CD= 4 -/2 em and AD= of x2- 4x.
c
B
A
x
<-------"'
28
D
em. Find the value
Singapore Mathematical Society
Singapore Mathematical Olympiad (SMO) (Senior Section Solutions)
2011
1. Ans wer: (B)
a2 b2 < Since a, b and c are nonzero real numbers and b2+c 2 c 2 +a 2 that b 2 +c 2 c 2 +a 2 a 2 +b 2 > > c2 a2 b2 Adding 1 throughout, we obtain
c2 , we s ee a 2+b 2
--=--
--=---
Thus
<
1 1 1 > > 2 , which implies that a 2 < b2 < c 2. So we have I a I a b2 c
2
<
l bl
2. Ans wer: (A) Note that (s in()+cos ())2= sin2 ()+cos 2()+2 s in() cos ()= 1+s in 2()= 1+a. Since 0::; ()::;� ' we have sin()+cos ()> 0. So s in() +cos ()= v'f+(i". 3. Ans wer: (D) We have � [(a- b)2 + (b- e)2 + (c- a)2 ] 2 [(-1)2 + (-1)2 +22 ] = 3. ·
�.
4. Ans wer: (C) 1 1 X 2y -
1 2x+y
2x+y =1 2y X y X 1 2+-- -- -= 1 X y 2 1 y X 2 X y 2x+y
::::?-
::::?-
::::?-
Now we have
30
-
<
l ei ·
Moreover, 2x
2(log 3+log 5� +log 7i) log 2 log(9 x 5)+log(49i) log(45 x 27 i) > ---,-log 2 log 2 log(45 x 3) log(128) = 7' > �g 2 �g2
------
SO X >
-
-
3. 5.
8. Ans wer: (B) Note that 74-1 = 2400, s o that 74n-1 is divis ible by 100 for any n E z+. Now, 756 7(756-1-1 +1) 7(756-1-1)+7 7(74n-1)+7 ,
where
56-1 '77+ E !LJ . 4 Since 7(74n-1) is divis ible by 100, its las t two digits are 00. It follows that the last two digits of 756 are 07. =
n
9. Ans wer: (A) logx 2011+logy 2011 logxy 2011
log 2011 log 2011 . log xy + ) ( ) log x logy log 2011 1 1 (- +-) . (log x +logy) log x logy log x logy 2+--+-logy log x 4 (us ing A M 2:: G M), (
>
and the equality is attained when log x = logy, or equivalently, x= y. 10. Ans wer: (C) The roots of the equation x2- (c- 1)x+c2- 7c+ 14 = 0 are a and b, which are real. Thus the dis criminant of the equation is non-negative. In other words , (c-1)2- 4(c2- 7c+ 14)= -3c2+26c- 55= (-3c+11)(c- 5) 2:: 0. 11 < < So we have c - 5. Together with the equalities 3 (a+b?- 2ab (c- 1)2-2(c2- 7c+ 14) -c2+12c- 27= 9 - (c- 6)2,
32
15. Ans wer: 8001 Note that J5n- y'5n- 4 < 0. 01 if and only if 4 > 400. y'5; + v'5n- 4= J5n 5n- y'5n- 4 If n= 8000, then J5n+y'5n- 4= v' 40000+\1'39996 < 400. If n= 8001, then J5n+y'5n 4= v' 40005+v'40001 > 400. So the ans wer is 8001. -
16. Ans wer: 1006 The s eries can be paired as 1 1 1 1 1 1 ) + + ( ) + ( + )+( + 1 11 · · · 11 1 +11- 1+ 111 · 1 + 11-20 1+11-2oo9 1+112oo9 1 +1120 Each pair of terms is of the form 1 1 ----,-+--= 1. 1 1+ a 1+a There are 1006 pairs of s uch terms , and thus the s um of the s eries is 1006. 17. Ans wer: 54 X
7 77r . 4 (-7r ) +cos 4 (-7r ) +s in 4 (8) +cos 4 (87r) 8 8
Sill
sin 4 (i) +cos 4 (i) + s in 4 (i) +cos 4 (i) . 2 (47r) = 3 . 2 (87r) + COS 2 (87r) )2- 4 Sill . 2 (87r) COS 2 (87r) = 2- Sill 2(Sill 2 Thus 36x = 54. 18. Ans wer: 1005 By the laws of s ine and cos ine, we have s in A s inE sinG and c b a Then 1 cos C cote cosAsi B + �os B sinA � sinC cot A+cot B smAsm B s in A s in B cos C sin(A+B) s inC s in A s in B ( . 2 ) cos C Sill 0 (ab jc 2) s in2 C a 2 +b2- 2 ) )( ( 2ab s in2 C a 2 +b 2- c 2 2c 2 2011- 1 = 1005. 2 .
34
·
22. Ans wer: 3 Note that 1 1 -+X y
1 =? xy- 211x- 211y 211
= -
=
0 =? (x- 211)(y- 211)
=
2112.
Since 211 is a prime number, the factors of 2112 are 1, 211, 2112, -1, -211, -2112. Thus the pairs of integers (x, y) s atis fying the las t equation are given by: (x- 211, y- 211)
=
(1, 2112), (211, 211), (2112 , 1), (-1, -2112), (-211, -211), (-2112, -1).
Equivalently, (x, y) are given by (212, 211 +2112), (422, 422), (211+2112 , 212), (210, 211- 2112), (0, 0), (211- 2112 , 210). Note that (0, 0) does not s atis fy the firs t equation. Among the remaining 5 pairs which s atis fy the first equation, three of them satis fy the inequality x � y, and they are given by (x, y) (422, 422), (211+2112 , 212), (210, 211- 2112). =
23. Ans wer: 93 By long divis ion, we have x 4 +ax 2+bx+c
=
(x2+3x- 1) · (x 2- 3x+(a+10))+(b-3a- 33)x+(c+a+10).
Let m1 , m2 be the two roots of the equation x2 + 3x- 1 0. Note that m1 -=1m2 , since the dis criminant of the above quadratic equation is 32-4 · 1 · 1· (-1) 13 -=10. Since m1, m2 als o s atis fy the equation x4 +ax 2+bx+c 0, it follows that m1 and m2 als o satis fy the equation (b- 3a- 33)x+(c +a+ 10) 0. =
=
=
=
Thus we have (b- 3a- 33)m1 +(c +a+ 10)
=
0,
and (b- 3a- 33)m2 + (c +a+10) 0. Since m1 -=1m2 , it follows that b-3a-33 0 and c+a+lO 0. Hence we have b 3a+33 and c -a-10. Thus a+b+4c+100 a+(3a+33)+4(-a-10)+100 93. =
=
=
=
=
=
=
24. Ans wer: 1120 Let m and n be pos itive integers s atis fying the given equation. Then 3(n2- m) 2011n. Since 2011 is a prime, 3 divides n. By letting n 3k, we have (3k)2 m + 2011k. This implies that k divides m. Let m rk. Then 9k2 rk+2011k s o that 9k r+2011. The smalles t pos itive integer r s uch that r+2011 is divis ible by 9 is r 5. Thus k (5+ 2011)/9 224. The corres ponding values of m and n are m 1120 and n 672. =
=
=
=
=
=
=
=
=
36
=
=
� = c2�b2• Since GN is parallel to AD and G is the centroid of the triangle ABC, we have MD/MN= 3. It follows that c+b= aJ3. Thus , AB = aJ3-b = 15-6= 9.
28. Ans wer: 28 The s um of 55 pos itive cons ecutive integers is at leas t (55 x 56)/2 = 1540. Let the middle number of thes e cons ecutive pos itive integers be x. Then the product abed= 55x = 5 11 x. So we have 55x 2': 1540 and thus x 2': 28. The leas t value of a+b +c +d is attained when x= 5(7). Thus the ans wer is 5+11+ 5+ 7 = 28. ·
·
29. Ans wer: 168 Firs t L_BD P= L_BCA= L_BAP s o that P, B , D, A are concyclic. Thus L_ACD = L_ PBA= L_ PDA= L_DAC s o that DA= DC.
By cos ine rule, cos C = 3/5. Thus DC = � AC jcos C = 21 x 5/3 = 35. Hence ED = 10 and BC = 10 + 35= 45. Thus area(L,ABD ) = �� x area(L,ABC). By Heron's formula, area(L,ABC)= 756. Thus area(L, ABD)= �� x 756= 168. 30. Ans wer: 276 Since the number of pos itive divis ors of a is odd, a mus t be a perfect square. As a is a divis or of 4400 = 24 x 52 x 11 and a has exactly 9 pos itive divis ors , we s ee that a= 22 x 52. Now the leas t common multiple of a and b is 4400 implies that b mus t have 24 x 11 as a divis or. Since 24 x 11 has exactly 10 pos itive divis ors , we deduce that b = 24 x 11 = 176. Hence a+b = 276. 31. Ans wer: 20 Firs t we let /!, be the line which extends BC in both directions . Let E be the point on /!, s uch that AE is perpendicular t o f. Similarly, we let F be the point on /!, s uch that DF is perpendicular to f. Then, it is eas y to s ee that BE = AE = v'6, CF = 2J2 and DF = 2v'6. Thus EF = y'6 + 4- 2J2 + 2J2 = 4 +v'6. Now we let G be the point on DF s uch that AG is parallel to £. Then AG = EF = 4 +v'6 and
38
which is a contradiction to the given fact that each element of S is les s than or equal to 2011. Case 2. 1 is in A. We may let a16 = 1. Then a 1, a 2 , As in Cas e 1, we have
· · ·
, a 1 5 are compos ite numbers .
which is a contradiction. Thus we have shown that every 16-element s ubs et A of S s uch that all elements in A are pairwis e relatively prime must contain a prime number. Hence the smallest k is 16. 35. Ans wer: 12 Let the extens ions of A Q and BP meet at the point R. Then LP R Q= LPAB= LQPR s o that QP = QR. Since QA = QP, the point Q is the midpoint of A R. As A R is parallel to LP, the triangles A RB and LP B are similar s o that M is the midpoint of PL. Therefore, N is the centroid of the triangle P LB, and 3 MN= BM.
B
Let LABP= e. Thus tanB= A R/AB= 2A QjAB= 5/6. Then B L= PB cos B= AB cos 2 0. Als o BM/BL = BQ/BA s o that 3MN = EM = �� AB cos 2 0 cos 2B( QM + 3MN). Solving 'for MN, we have MN= 3 �:;_; e = 3x (�f6)2 = 12.
40
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO)
2011
(Senior Section, Round 2 solutions)
1. There is an error in this problem. The triangle is not neces s arily equilateral. In fact
we s hall prove that the altitude at A, the bis ector of LB and the median at C meet at a common point if and only if cos B= � where BC= a, CA = b and A B= c. a
c
Let D,E and F be the points on BC,CA and A B res pectively s uch that AD is the altitude at A , BE is the bis ector of LB and CF is the median at C. Suppos e that AD,BE,CF meet at a common point. The point of concurrence of AD,BE and CF mus t lie ins ide the triangle A BC. Since F is the midpoint of A B, by Ceva's theorem CE : EA = CD : DB. Us ing the angle bis ector theorem, CE : EA =a : c. Thus CD= a2/(a+c) and DB= acj(a+c). Thus cos B= �� = � a
c·
A
B
Convers ely, if cosB= � then LB is acute and BD= c cos B= acj(a+c)
' c
So given a and c, the acute angle B and hence the triangle A BC is determined. If a -=/- c, then the triangle A BC is not equilateral. 2. Yes , in fact, for any k E
N,
there is a s et 8k having k elements s atis fying the given condition. For k = 1, let 81 be any s ingleton s et. For k = 2, let 82 = {2 ,3}. Suppos e that 8k = {a1, ... , ak} s atis fies the given conditions. Let b1 = a1a2 ak bj = b1+aj -1, 2 :::; j :::; k+ 1. · · ·
Let 8k+1 = {b1, b2, ... , bk+1}. Then the fact that 8k+1 s atis fies the required property can be verified by obs erving that lm- nl = (m, n) if and only if (m- n) divides m. 42
Note that a1 a2 ··· an
=
1. It suffices to show that
since it is equivalent to 1 1 1 �n- 1. + +···+ 1+an 1 +a1 1+a2
--
We shall show that ( * ) is true for n 2:: 2. The case n 2 is obvious. We will now prove it by induction. Suppose ( * ) holds for n k. Now assume a1 ··· ak+l 1, ai rel="nofollow"> 0 for all i. To prove the inductive step, it suffices to show that =
=
which can be verified directly. Note: This is an extension of the problem :
44
=
Throughout this paper, let lxJ denote the greatest integer less than or equal to x. For example, l2.1J = 2, l3.9j = 3 ( This notation is used in Questions 7, 9, 19 and 20). 1.
A circular coin A is rolled, without sliding, along the circumference of another stationary circular coin B with radius twice the radius of coin A. Let x be the number of· degrees that the coin A makes around its centre until it first returns to its initial position. Find the value of x.
2. Three towns X,Y and Z lie on a plane with coordinates (0, 0), (200, 0) and (0, 300) respec tively. There are 100, 200 and 300 students in towns X,Y and Z respectively. A school is to be built on a grid point (x, y) , where x and y are both integers, such that the overall distance travelled by all the students is minimized. Find the value of x+y. 3. Find the last non-zero digit in 30!.
( For example, 5! = 120; the last non-zero digit is 2.) 4. The diagram below shows �A BC , which is isoceles with A B = AC and LA 20°. The point D lies on AC such that A D = BC. The segment E D is constructed as shown. Determine LA B D in degrees. A =
cos4 a sin4 a cos4 f3 sin 4 f3 . 5. G1ven that �(3 + -.-2- = 1, evaluate --2- + -.-2 - . cos sm f3 cos a sm a 6. The number 25 is expressed as the sum of positive integers XI, x 2, , Xk, where k What is the maximum value of the product of XI, x 2, X3, , and Xk ? ·
·
46
·
·
·
·
::;
25.
18.
A collection of 2011 circles divide the plane into N regions in such a way that any pair of
circles intersects at two points and no point lies on three circles. Find the last four digits of N.
19. If a positive integer N can be expressed as lxJ + l2xJ + l3xJ for some real numbers x, then we say that N is ''visible"; otherwise, we say that N is "invisible". For example, 8 is visible since 8= l1.5J + l2 (1.5 )J + l3 ( 1.5 )J , whereas 10 is invisible. If we arrange all the "invisible" positive integers in increasing order, find the 2011th "invisible" integer. 20. Let A be the sum of all non-negative integers
n
satisfying
Determine A. 21.
A triangle whose angles are A, B, C satisfies the following conditions sin A+ sinE + sinG cos A + cos B + cosC and
12 7'
12 . 25 Given that sin C takes on three possible values 81, 82 and 83 , find the value of 100818283 . sin A sin B sin C =
22. Let x
>
1,
y
>
1 and
z >
1 be positive integers for which the following equation 1! + 2! + 3! + . . . + x!=
y
z
is satisfied. Find the largest possible value of x + y+ z. 23. Let A BC be a non-isosceles acute-angled triangle with circumcentre 0, orthocentre H and LC = 41 o. Suppose the bisector of LA passes through the midpoint M of OH. Find LHAO in degrees. 24. The circle 'Yl centred at 01 intersects the circle 'Y2 centred at 02 at two points P and Q. The tangent to 'Y2 at P intersects 'Yl at the point A and the tangent to 'Yl at P intersects 'Y2at the point B where A and B are distinct from P. Suppose PQ · 0102= P01 P02 and LAP B is acute. Determine the size of LAP B in degrees. ·
25. Determine
n
1 . L (n). -+oo . n lim
i=O
( Note: Here
(�) 't
denotes
n�
.1 't. n
(
" ' 't .
)
�
fori= 0, 1, 2, 3, · · ·
48
,
n.
)
Thus, 30! 30! 107
1 226 . 3 4 . 5 7 . 74 . 112 . 132 . 17 . 19 . 23 . 29 1 1 2 9 . 3 4 . 7 4 . 112 . 132 . 17 . 19 . 23 . 29 1 6 4 . 25 . 74 . 112 . 132 . 17 . 19 . 23 . 29 6 (2) (1) (1) (9) (7) (9) (3) (9) (mod 10) 2 (-1) (-3) (-1) (3) (-1) (mod 10) 8(mod 10),
showing that the last non-zero digit is 8. D
4. Answer.
10
Solution.
Let E be the point inside f:,A BC such that f:,E BC is equilateral. Connect A and D to E respectively. It is clear that f:,AE B and f:,AEC are congruent, since AE = AE , A B = AC and BE= CE. It implies that L BAE = LCAE= lOa . Since A D = BC = BE , LE BA = L D A B = 20a and A B = BA, we have t:,A BE and f:, BA D are congruent, implying that LA B �= L BAE = lOa .
D
5.
Answer.
.
Solutwn.
1 cos2 a sin2 a cos4 a sin4 a . (3 --:--- . Then 13 and sine= ---Smce �(3 + . 2 (3 = 1, set cose= cos cos sm sm c
---
---
cos (e- a )= cose cos a + sine sin a= cos2 a + sin2 a= 1. 50
Now let x2 +ax+ b= (x- xl) (x- x2), where x1:::; x2. Then the set of integer solutions of x2+ ax+ b< 0 is {k : k is an integer, x1 < k< x2}. By the given condition, {k : k is an integer, x1 < k< x2}= {k : k is an integer, -11< k< 6} = {-10,-9, . .· , 5}. Thus -11 :::; x1 < -10 and 5 < x2 :::; 6. It implies that -6 < x1 + x2 < -4 and -66 :::; X1X2< -50. From x2+ax+b= (x-x1) (x-x2), we have a= - (x1+x2) and b= x1x2. Thus 4< a< 6 and -66:::; b<-50. It follows that 54< a - b< 72. Thus
max:Lia- biJ :::; 71.
It remains to show that it is possible that lla - biJ = 71 for some a and b. Let a= 5 and b= -66. Then x2+ab+b= (x+11) (x-6) and the inequality x2+ab+b< 0 has solutions {x: -11< x< 6}. So the set of integer solutions of x2 + ab + b< 0 is really the set of integers in A n B. Hence 10.
max:Lia- biJ = 71.
Answer. Solution.
D
8 We consider the polynomial P(t)= t3 + bt2 + ct+ d.
Suppose the root of the equation P (t)= 0 are x, y and z. Then
-b= X+y+ Z= 14,
and x3 +y3 + z3 +3d= (x+ y+ z) (x 2 +y2+ z2- xy- xz- yz). Solving for b, c and d, we get b= -14, c= 30 and d= -64. Finally, since t3 - 14t2+30tD 64= 0 implies t= 2 or t= 4 or t 8, we conclude that max{ a, ,B, 'Y}= 8. =
11.
Answer.
38
Solution. Let n be an even positive integer. Then each of the following expresses n as the sum of two odd integers: n= ( n- 15) + 15, ( n - 25) + 25 or ( n - 35) +35. Note that at least one of n - 15, n - 25, n - 35 is divisible by 3, hence n can be expressed as the sum of two composite odd numbers if n > 38. Indeed, it can be verified that 38 cannot be expressed as the sum of two composite odd positive integers. D
12.
Answer.
1936 52
We shall show that I AI bijection from A to B:
=
I BI
(63°)
=
=
34220 by showing that the mapping¢> below is a
First, since { a1, a2, a 3 } E A, we have a1+3 implying that {a1, a2- 2, a 3 - 5} E B.
�
a2 and a2+4
�
a 3 , and so a1
<
b3 , we have {b1, b2+2, b3+5}
<
a2-2
<
a 3-5,
It is clear that¢> is injective. It is also surjective, as for any {b1, b2, b3 } A and Hence¢> is a bijection and I AI 16.
Answer.
=
I BI
=
E
B with b1
<
b2
34220.
E
0
32
It is clear that 8(cos 40°) 3 -6 cos 40°+ 1 = 0, since cos 3A = 4 cos3 A-3 cos A . Observe that Solution.
1 3 + 64 sin2 20° 2 2 sin 20° cos 20° 2 6 + 32 (1- cos 40°) 1- cos 40° 1 + cos 40° 8 cos 40°+ 4 + 32 32 cos 400 1- (cos 40°) 2 8 COS 40° + 4- 32 COS 40° + 32 ( COS 40°) 3 + 32 1- (cos 40°) 2 1- 6 cos 40°+8(cos 40°) 3 4x +32 1- (cos 40°)2 32, ----
_
==
where the last step follows from 8(cos 40°) 3 - 6 cos 40° + 1 = 0. 17.
Answer. Solution.
0
6029 Given the original equation f(x2 + x) + 2f(x2-3x + 2) = 9x2- 15x,
we replace x by 1-x and obtain f(x2-3x+ 2) + 2f(x2+ x) = 9 (1-x) 2- 15 (1-x) = 9x2-3x-6. Eliminating f(x2-3x+ 2) from the two equations, we obtain 3f(x2 + x) = 9x2+ 9x- 12, thereby
f (x2+ x) = 3x2 + 3x-4 = 3 (x2 + x)- 4,
hence f (2011) = 3(2011)- 4 = 6029. 0
54
20.
Answer.
95004
Solution. We shall prove that for any positive integer nonnegative integer solutions to L�J= La�lJ, then
a, if f(a) denotes the sum of all
f(a)= 16a(a2-1)(a+ 2).
Thus
!(27)= 95004.
Let n be a solution to L�J = La�1J . Write n= aq + r, where 0::; r
A= LL(qa+r) q=O r=q a-1
a-1a-1
= L(a-q)qa+ LLr = = =
q=O r=q a-1
q=O a-1
a-1
q=O a-1
q=O a-1
r=O q=O a-1
q=O a-1
q=O a-1
r=O a-1
a-1
q=O
q=O
r=O
r=O
r
a2L:q-aL q2+ LLr a2L:q-aL q2+ L:r(r+1) a2L q-aL q2+ Lr2+ Lr
1) (a2+ 1 ) · �a( a-1 ) + (1- a)· �a(2a-1)(a2 6 1 = 6a(a2-1)(a+ 2). =
D
21.
Answer.
48
By using factor formulae and double angle formulae: sin A + sinE + sinG 4cos .:1 cos fl. cos Q 12 2 2 2 - --cosA + cos B + cosC 7' 1+4sin 4 sin � sin � and
A B C 12 . . . . A . B . C smA sm B smC = 8 sm 2 sm 2 sm 2 cos 2 cos 2 cos 2= .
25
Solving these equations, we obtain . A . B . C sm-sm-sm- - 0 .1
2
2
2
B C A cos 2 cos 2 cos 2 Furthermore, sin
C
2 = cos
(
A+ B
2
)
= cos 56
A
0. 6
B
A
B
2 cos 2 - sin 2 sin 2'
p
L
be the midpoint of BC . It is a known fact that AH = 20£. To see this, extend meeting the circumcircle of the triangle A BC at the point N. Then AN BH is a parallelogram. Thus AH = NB = 20£. Therefore in the right-angled triangle OLC, OC = OA = AH = 20£. This implies L.OCL = 30°. Since the triangle A BC is acute, the circumcentre 0 lies inside the triangle. In fact L.A = 60° and L.B = 79°. Then L.OAC = L.OCA = 41°-30° = 11°. Consequently, L.HAO = 2L.OAM = 2 X (30°- 11°) = 38°. D Let
CO
24.
Answer.
30
Let P01 = r1 and P02 = r2. First note that 0102 intersects PQ at the midpoint H (not shown in the figure) of PQ perpendicularly. Next observe that L.APQ = L.PB Q = L.P0201, and L.BPQ = L.PA Q = L.P0102. Therefore L.APB = L.APQ + L.BPQ = L.P0201 + L.P0102.
'/'1
'/'2
H and sin f3 = Let L.P0201 = a and L.P01 02 = f3. Then sin a = PrQ, cos a = 02 r 2 2 2 fr�, cos f3 = 0f1H. Thus sin L.APB = sin ( a + /3) = sin a cos f3+ cos a sin f3 = fr� · 0f1H +
25.
0r2H. E9.. 2 2r1
=
Answer.
2
Solution.
PQ(· 01 H+02H) = PQ0 · 102 2r1r2 = 21. Since L.APB is acute it is equal to 30°. 2r1r2 '
Let
Assume that
n
�
3. It is clear that
() � ; -1
an=
2+
n
�=1
58
-1 rel="nofollow">
2.
·
D
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO)
2011
(Open Section, Round 2) Saturday, 2 July 2011
0900-1330
1. In the acute-angled non-isosceles triangle ABC, 0 is its circumcentre, His its orthocentre and AB > AC. Let Q be a point on AC such that the extension of HQ meets the extension of BC at the point P. Suppose BD= DP, where Dis the foot of the perpendicular from A onto BC. Prove that LODQ = 90°. 2. If 46 squares are colored red in a 9 x 9 board, show that there is a 2 x 2 block on the board in which at least 3 of the squares are colored red. 3. Let x, y, z > 0 such that
� +� + �
<
x!z·
Show that
2z 2x 2y + + yl + x2 y'l + y2 yl + z2 4.
Find all polynomials P(x) with real coefficients such that P(a)
5.
<
E Z
implies that
a E Z.
Find all pairs of positive integers (m, n) such that m+n-
3mn m+n
60
2011 3
3.
which leads to a contradiction. On the other hand, s uppos e that r1, r3, r5, r7, r9 2:: 6. Then the s um of any 2 cons ecutive ri's is :::; 9. Again we get a contradiction as (r1+r2)+
·
·
·
+(r7+rs)+rg
:::; 4
9+9=45.
x
Let r = 1/x, s= 1/y, t= 1/z. There exis ts a < 1 s uch that r + s + t=a2rst or a(r + s + t)=a3rst. Let a=ar, b=as, c=at. Write a= tan A, b tanB, c = tan C, then A+ B + C 1r. It is clear that 3.
=
=
1 2
- X
1 1 1 + ----= + -;= ::;;;: y1+r2 v1+s2 v1+t2 1 1 1 < + ----= + -;= ::;;;: v1+a2 v1 + b2 v1+c2 = cos A+ cos B+ cos B A+ B + C < 3 cos =�=� x RHS. 2 2 3
LHS=
(
)
2nd soln: Note that
1 1 1 -+ - + x y z
Hence
2x =;;;= y1+x2
<
1 xyz
:::::?
xy+yz+xz
2x = == Jx2+xy+xz+yz
----r=
< -r-::::::==
=
By AM-GM we have
==
<
1.
2x J(x+y)(x + z)
X 2x X -=;= < . + J(x+y)(x+z) - x+y x+z
----r-.;======:=;=
= --
Similarly, z z 2z =;= < -- + --. V( Z+X) (z + y) Z+X Z + y
2y Y + __ Y < __ J(y + z)(y+x) - y + z y + x'
----r-.;======:=;=
=
The des ired inequality then follows by adding up the three inequalities. 4.
Let P(x)=anxn + +a1x+a0. Define Q(x)=P(x + 1) - P(x). Then Q(x) is of degree n- 1. We'll prove by contradiction that IQ(x)l :::; 3 for all x. This will imply that n :::; 1. As s ume that IQ(a)i > 3 for s ome a E R Then IP(a + 1) - P(a)l > 3. Thus there are 3 integers between P(a) and P(a + 1). Hence there exis ts three values ·
·
·
62
Singapore Mathematical Society . Singapore Mathematical Olympiad (SMO)
2010
(Junior Section)
Tuesday, 1 June 20 10
0930-1200
Important:
Answer ALL 35 questions. Enter your answers on the answer sheet provided. For the multiple choice questions, enter your answer on the answer sheet by shading the bubble containing the letter (A, B, C, D or E) corresponding to the correct answer. For the other short questions, write your answer on the answer sheet and shade the appropriate bubbles below your answer. No steps are needed to justify your answers. Each question carries 1 mark. No calculators are allowed.
PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO.
1
Multiple Choice Questions
1. Among the five real numbers below, which one is the smallest? (A)
v2010;
'JJYJ_J�
(B)
v2009;
(C) 2010;
20!.0MIV\n
(D)
2010 ; 2009
(E)
2009 . 2010
2. Among the five integers below, which one is the largest?
3. Among the four statements on real numbers below, how many of them are correct? "If a
0 then !b < ! a
"If a
"· '
"If a
(A) 0;
4.
(B)
(C) 2;
1;
(D)3;
(E) 4.
What is the largest integer less than or equal to (A) 2009;
(B)
2010;
(C) 2011;
(D) 2012;
.{/(2010)3 +3 x (2010)2+4 x 2010+1? (E) None of the above.
5. The conditions of the road between Town A and Town B can be classified as up slope, horizontal or down slope and total length of each type of road is the same. A cyclist travels from Town A to Town B with uniform speeds 8 km/h, 12 km/h and 24 km/h on the up slope, horizontal and down slope road respectively. What is the average speed of his journey?
12kmfh ----+
Town A
(A) 12 km/h;
(B)� km/h;
Town B
(C)
16 km/h; 2
(D) 17 km/h;
(E) 18 km/h.
6. In the diagram, MBC and f).CDE are equilateral triangles. Given that LEBD LAEB X0, what is the value of x?
=
62° and
=
A
D (A) 100;
(B) 118;
(C) 120;
(D) 122;
(E) 135.
7. A carpenter wishes to cut a wooden 3 x 3 x 3 cube into twenty seven 1 x 1 x 1 cubes. He can do this easily by making 6 cuts through the cube, keeping the pieces together in the cube shape as shown:
. . . . :·
.
.
:
·······.·······.·······
What is the minimum number of cuts needed if he is allowed to rearrange the pieces after each cut? (A) 2;
(B) 3;
(C) 4;
(D) 5;
(E) 6.
8. What is the last digit of 7(77)? (A) 1;
(B) 3;
(C) 5;
(D) 7;
(E) 9.
3
9.
Given that n is an odd integer less than many such integers are there ? (A) 3;
10.
(B)
4;
(C) 5;
(D) 6;
1000 and the product of all its digits is 252.
How
(E) 7.
What is the value of
(A) 451856;
(B)
691962;
(C) 903712;
(D) 1276392;
(E) 1576392.
Short Questions
11.
Let x and y be real numbers satisfying y=
2008x+2009 + 2010x- 2011
2008x+2009 +2010· 2011- 2010x
Find the value of y. For integers ah . . ., an
12.
E {1, 2, 3, . .., 9}, we use the notation a1a •••an to denote the number 2 10n-1a1+10n-2a2+· · · +10an-1+an. For example, when a = 2 and b = 0, ab denotes the number 20. Given that ab = b2 and ache = (ba)2• Find the value of abc.
13.
Given that (m- 2) is a positive integer and it is also a factor of 3m2- 2m+10. Find the sum of all such values of m.
14.
In triangle ABC, AB = 32 em, AC = find the length of AM in em.
36 em and BC = 44 em.
A
B
M
4
c
If M is the midpoint of BC,
15. Evaluate
678+690+702+ 714+...+ 1998+2010 3+9+15+21+... + 327+ 333
16. Esther and Frida are supposed to fill a rectangular array of 16 columns and 10 rows, with the numbers 1 to 160. Esther chose to do it row-wise so that the first row is numbered 1, 2, ..., 16 and the second row is 17, 18, . .., 32 and so on. Frida chose to do it column wise, so that her first column has 1, 2, . .., 10, and the second column has 1 1, 12, ..., 20 and so on. Comparing Esther's array with Frida's array, we notice that some numbers occupy the same position. Find the sum of the numbers in these positions.
... .. . ... ... ... ... ... ... 145 146 147 ... 2 18 ... .
1 17
.
3 19
.
... 16 . . . 32 . . . .. . . ... .. . 160 .
.
1 2 .. . .. .
11 12 ... . ..
10
20
.. ... . .. .. . 30 ...
21 22 ... ...
.
... ... .. . ... ...
151 152 .. .. .
.
160
Frida
Esther
17. The sum of two integers A and B is 2010. If the lowest common multiple of A and B is 14807, write down the larger of the two integers A or B.
18. A sequence of polynomials an(x) are defined recursively by a0(x) a1 (x) an(X)
=
=
=
1, x2+x+1, (� + 1)an-I(X)- an_z(X), for all n 2 2.
For example,
az(x) a3(x)
=
=
=
(�+1)(�+x+ 1)- 1 x4+x3 +2x2+x, (� +1)(x4+x3 + 2�+x) (� +x+1) x7 +x6+2x5 +2x4 + x3 +�- 1. =
-
Evaluate a 010(1). 2 19. A triangle ABC is inscribed in a semicircle of radius 5. If AB of s2 where s AC+BC. =
5
=
10, find the maximum value
102009>.
20.
Find the last two digits of 2011<20
21.
Your national football coach brought a squad of 18 players to the 2010 World Cup, consisting of 3 goalkeepers, 5 defenders, 5 midfielders and 5 strikers. Midfielders are versatile enough to play as both defenders and midfielders, while the other players can only play in their designated positions. How many possible teams of 1 goalkeeper, 4 defenders, 4 midfielders and 2 strikers can the coach field?
22. 23.
Given that value of x.
169(157-77x)2 +100(201-100x)2 = 26(77x-157)(1000x-2010),
Evaluate
find the
(20202-20100)(201002-1002)(20002+20100) 20106-106
24.
When 15 is added to a number x, it becomes a square number. When 74 is subtracted from x, the result is again a square number. Find the number x.
25.
Given that x and y are positive integers such that 56::::; x+y::::;
26.
the value of f-x2.
59 and 0.9
<
:: y
<
0.91, find
Let AA' and BB' be two line segments which are perpendicular to A' B'. The lengths of AA', BB' and A' B' are 680, 2000 and 2010 respectively. Find the minimal length of AX+XB where X is a point between A' and B'.
B
I
0 0 0 C'l
0 00 1,0
l
A'
27.
The product 1 x 2 x 3 x M= 1! x 2! x 3! x 4! x 5! ·
·
·
2010
n is denoted by n!. For example 4! = 1 x 2 x 3 x 4 = 24. Let 6! x 7! x 8! X 9!. How many factors of Mare perfect squares?
x
x
1
B'
6
28. Starting from any of the L's, the word LEVEL can be spelled by moving either up, down, left or right to an adjacent letter. If the same letter may be used twice in each spell, how many different ways are there to spell the word LEVEL?
L
L
L
E
L
E
v
E
L
E
L
L
'
L ...___
29. Let ABCD be a rectangle with AB 10. Draw circles C1 and C2 with diameters AB and CD respectively. Let P, Q be the intersection points of C1 and C2. If the circle with diameter PQ is tangent to AB and CD, then what is the area of the shaded region? =
30. Find the least prime factor of
10000 .. ·00 1. ""---v-"
2010-many
7
�
�
31. Consider the identity 1 + 2 +· · · + n = n(n+ 1). If we set P1(x) = x(x + 1), then it is the unique polynomial such that for all positive integer n, P1 (n)=1 + 2 +· · · + n. In general, for each positive integer k, there is a unique polynomial P (x) such that k P (n) =1k + 2k + 3 k + ··· + nk k Find the value of P2o10C
-
�)
for each n=1, 2, . . . .
.
32. Given that ABCD is a square. Points E and F lie on the side BC and CD respectively. such that BE=CF = AB. G is the intersection of BF and DE. If
�
Area of ABGD m ----- = Area of ABCD n is in its lowest term, find the value of m + n.
A
.-------�
B E
F
C
33. It is known that there is only one pair of positive integers a and b such that a :::; b and a2 + b2 + 8ab=2010. Find the value of a + b. 34. The digits of the number 123456789 can be rearranged to form a number that is divisible by 11. For example, 123475869, 459267831 and 987453126. How many such numbers are there? 35. Suppose the three sides of a triangular field are all integers, and its area equals the perimeter (in numbers). What is the largest possible area of the field?
8
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2010 (Junior Section Solutions) 1. Ans: (E) It is the only number less than 1. 2. Ans: (D) Other than (D), all numbers are less than 20102010. Now 37 9 3(3 ), the result follows.
>
2010. Thus 20102010
<
(37l
<
3. Ans: (B) Only the third statement is correct: a < b implies a+ c < b + c. For other statements, counterexamples can be taken as a= -1, b= 1; c= 0 and a= 0, b= -1 respectively.
4. Ans: (C) Since (2010+ 1)3= 20103 + 3 x 20102+ 3 x 2010+ 1. The result follows.
5. Ans: (A)
Let the distance between Town A and Town B be 3s. The total time taken for up slope, horizontal and down slope road are �, { and ;4 respectively. His average speed for the 2 3s . 1 . km/h. Wh 0 e JOUrney IS s s s = 12 8 + 2 + 24 1
6. Ans: (D)Observe that !:l.BCD is congruent to !:l.ACE (using SAS, BC= AC, CD= CE and LACE= 60° - LECB= LBCD). Thus LAEC= LBDC. We get x
o
= = = =
360° - LAEC- LBEC 360° - LBDC- LBEC
LEBD+ LECD 62° + 60°= 122°
Thus x= 122.
9
A
D
7.
Ans: (E)
There is no way to reduce the cuts to fewer than 6: Just consider the middle cube (the one which has no exposed surfaces in the beginning), each of the its sides requires at least one cut.
8. Ans: (B)
The last digit of 7k is 1, 7, 9, 3 respectively for k= 0, 1, 2, 3 (mod Since 77 = (-1? = 3 (m�d
9.
Ans: (C)
252 = 2 x 2 x 3
10.
x
3
x
4), the last digit of 7 <77) is 3.
4).
7. We can have: 667, 497, 479, 947, 749.
Ans: (C) We only need a rough estimate to rule out the wrong answers. -fiT� 3.3 and -Y5� 2.2, so the sum is� 5 . 58 + (1.1)8 � 5. 58 = 30.254 � 304 � 810000. Thus (C). Of course the exact answer can be obtained by calculations, for example,
t (�)< «t")l( �)8-1 t (�)< «t")'(- �)8-1 +
=2
.I (�)( m)i( vs)8-i
zeven
10
1 1. Ans: 20 10. Let
a=
2008x+ 2009 20 10x- 20 11·
Then a � 0 and- a � 0 since they are under the square root. Hence a= 0. Thus y= 20 10.
12. Ans: 369.
It is easy to see that b= 5, ab= 25 or b= 6, ab= 36. Upon checking, b = 6, a= 3 and so 2 (b a?= 63 = 3969. Therefore c = 9. Hence abc= 369.
13. Ans: 5 1. Since
2 3m - 2m+ 10
18 m- 2 m- 2 is an integer. Thus m- 2 is a factor of 18. m- 2= 1, 2, 3, 6, 9 , 18, thus m = 3, 4, 5, 8, 11, 20. The required sum is 5 1. -----
14. Ans: 26. Using the Median Formula,A_MZ
=
= 3m+ 4+
2 2 2 2AB + 2AC - BC ·
4
--
. ThusA M= 26 em.
15. Ans: 16. Note that both numerator and denominator are arithmetic progressions. Removing common factors gives
2X
113+ 115+ 117+ ... + 333 + 335 1+ 3+ 5 + ...+ 109+ 111
2 2 X 112 (335 + 1 13) = 5; (111+ 1)
:::::
16.
16. Ans: 322. The number in the r-th row and c-th column has value 16(r- 1) + c in Esther's array and value 10(c- 1) + r in Frida's array. So we need to solve
16(r- 1) + c= 10(c- 1)+ r
�
5r= 3c + 2.
There are exactly four solutions
(r, c)= {(1, 1), (4, 6), (7, 11), (10, 16)}.
11
17. Ans: 1139. A direct way to solve the problem is to factor 14807 directly. Alternatively, one may hope for A and B to have common factors to simplify the problem. This is a good strategy because of the following fact: "The greatest common divisor of A and B, equals the greatest common divisor of A + B and lcm(A, B). "
2010 is easily factored as 2 X 3 x 5 x 67. Checking that 67 is also a factor of 14807, we can conclude that 67 is also a factor of A and B. The problem is reduced to finding a and b such that 2010 1 07 =30 and ab= a + b= =221. 67 7 Since 221 can be factored easily, a andb must be 13 and 17. So the answer is 17x67=1139.
�
18. Ans: 4021. an(1) is the simple recurrence relation fn = 2fn-1 - fn- • fo = 1 and !1 = 3. Using standard 2 technique or simply guess and verify that fn=2n + 1. So a oio(1)=f2010 = 2(2010) + 1. 2 19. Ans: 200. ABC must be a right-angled triangle. Let x = AC and y = BC , by Pythagoras theorem r + r =102•
i =(x + y)2 =r
+ y2 + 2xy=100 + 2 x
area of ABC.
Maximum area occurs when x=y, i.e. LC AB=45°. So x=y= -{50.
20. Ans: 01. Note that 2011 = 11 (mod 100) and 112 = 21,11 3 (mod 100). Since 20102009 is divisible by 10,
20112 0102009
=
1110 x
·
·
·
x 1110
=
=
31 (mod 100) etc. So 1110
1
_
1
(mod 100).
21. Ans: 2250.
(i) x (;) x (�) choices for goalkeepers, strikers and midfielders respectively. The remaining
midfielder and defenders can all play as defenders, hence total number of possibilities are
3
X
lOX 5
12
X
(�)
=2250.
22. Ans: 31.
Let a= 1001x '- 2041 and b= 1000x- 2010.
Then the equation becomes a2+b2= 2ab. Thus (a- b?= 0. The result follows. 23. Ans: 100. Let x= 2010 and y= 10. The numerator becomes [(x+ y)2- xy] (�i ·
-
ii )
·
[(x - y)2+ xy]
= (� + xy + i ) iCx- y)(x+ y) (� - xy + i ) = Y2(� -l)(x3 +l) = y2(x6 -l). ·
·
Hence the answer is 100. 24. Ans: 2010. Let 15+x= m 2 and x-74= n2• We havem 2- n2= 89= 1x89. (m-n)(m+n)= 1x89. Let m-n= 1 and m+n= 89. Solving m = 45 and n= 44. Thus the number x is 452-15= 2010. 25. Ans: 177. From 0.9y < x < 0.91y, we get 0.9y + y < x + y < 0.91y + y. Thus 0.9y + y < 59 and 0.91y + y> 56. It follows that y < 31.05 and y> 29.3. Thus y= 30 or 31. If y= 30, then 27 < x < 27.3, no integer value of x. If y= 31, then 27.9 < x < 28.21, thus x= 28. Thus y2 - x2= (31 + 28)(31- 28)= 177. 26. Ans: 3350. Take the reflection with respect to A' B'. Let A" be the image of A. Then the minimal length is equal to the length of A" B = 20102 + (2000+ 680)2= 3350.
.V
27. Ans: 672. M
= l!x2!x3!x4!x5!x6!x7!x8!x9! = 28 X 37 X 46 X 55 X 64 X 73 X 82 X 9 = 230 X 313 X 55 X 73
A perfect square factor of M must be of the form 22xx32Yx52zx72w, where x, y, z and ware whole numbers such that 2x � 30, 2y � 13, 2z � 5, 2w � 3. Hence, the number of perfect square factors of M is 16x7x3x2= 672. 13
28. Ans: 144. There are total 12 ways starting from any of the L's to reach the middle V. Hence the total number of ways to spell the word LEVEL is 122 = 144. 29. Ans: 25. Let N be the midpoint of CD. Then LPNQ = 90°. So PQ = 5 Vi.
B
A
D
N
c
Then the area of the shaded region is
30. Ans: 11. Clearly 11 is a factor and 2, 3, 5 are not. We only need to rule out 7. 102011+1 = 4 (mod 7) because 103 = -1 (mod 7). 31. Ans: 0. Let k be a positive even numb�r.
14
k Define j(x) = Pk(x) - Pk(x- 1). Then j(n) = n for all integer n � 2. Note thatj is a . polynomial. We must have j(x) = xk. In particular, for integers n � 2, Pk(-n+ 1)- Pk(-n)
=
Pk(O)- Pk(-1)
=
j(-n+ 1)
=
(n- 1)\
Pk(-n+ 2)- Pk(-n+ 1) = j(-n+ 2) = (n- 2i,
Pk(1)- Pk(O)
k O , k j(1) = 1 . j(O)
=
=
k k k Summing these equalities, Pk(l)- Pk(-n) = 1 + O + 1 +
Define g(x) = Pk(-x) + Pk(x - 1). Then g(n) polynomial, g(x) = 0. In particular, Pk(
-�)+ Pk( -�)
32. Ans: 23. Join BD and CG and note that
=
0, i.e., Pk( -�)
��
=
A
=
=
·
·
·
+ (n- 1)k. That is,
0 for all integers n � 2. Since g is a
0.
2.
....-----.
B E
F
Assume the length of AB is 1. Let the area of ABGE and AFGC be x andy respectively.
Then the areas of t1EGC and ADGF are 2x arid 2y. Since the area of ABFC is
�
· we have
1 1 2 1 . . 3x+ y= 6. Smnlarly, 3y+ 2x=the area of AVEC= 3. Solve x = andy= . Thus 42 21 Area of ABGD Area of ABCD
=1
So m= 9 and n= 14. The result follows.
15
_
1 3( x + y)=l _ 5 42
=
�· 14
33. Ans: 42. Since a � 1,20 10 =a2+b2+8ab � 1+b2+8b. b2+8b-2009 � 0. However b2+8b-2009=0 has an integer solution 41. So a= 1 and b=41. The result follows. 34. Ans: 3 1680. Let X and Y be the sum of the digits at even and odd positions respectively. Note that 1 + 2 + 3 + + 9=45. We have X+ Y=45 and 11 divides IX- Yl. It's easy to see X= 17 and Y = 28; or X = 28 and Y = 17. Hence we split the digits into 2 sets whose sum is 17 and 28 respectively. ·
·
·
There are 9 ways for 4 digits to sum to 17: {9,5,2, 1}, {9,4,3, 1}, {8,6 ,2, 1}, {8,5,3, 1}, {8,4,3,2}, {7,6,3, 1}, {7,5,4, 1}, {7,5,3,2}, {6,5,4,2}. There are 2 ways for 4 digits to sum to 28: {9,8,7,4}, {9,8,6,5}. Thus the total number of ways is 11 x 4! X 5! =3 1680. 35. Ans. 60. Let the three sides of the triangle be a,b,c respectively. Then
where s=
�
a+ + c
�s(s - a)(s - b)(s - c)=a+ b + c=2s, . Note that s is an integer; otherwise s(s-a)(s-b)(s-a) is a non-integer.
Let x=s - a, y= s - b and z=s - c. Then x,y,z are positive integers satisfying xyz=4(x + y + z). Assume that x � y � z. Then xyz � 12x, i.e. , yz � 12, and thus z � 3. If z = 1, xy = 4(x + y + 1) implies (x - 4)(y - 4) = 20 = 20 (x,y)=(24,5),( 14,6),(9,8).
·
1
= 10 2 = 5 4. So ·
·
If z=2, 2xy=4(x+ y+ 2) implies (x- 2)(y- 2)=8=8· 1=4 2. So (x,y) =( 10 ,3 ),(6,4). ·
If z=3, 3xy=4(x + y + 3) implies (3x - 4)(3y - 4)=52, which has no solution x � y � 3. The area is 60, 42, 36, 30 , 24, respectively; and the largest possible value is 60.
16
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO)
2010 -'
(Junior Section, Round 2) Saturday, 25 June 2010
0930-1230
INSTRUCTIONS TO CONTESTANTS
1.
2. 3.
Answer ALL 5 questions. Show all the steps in your working. Each question carries
10
mark.
4. No calculators are allowed.
ABC D intersect at S and let P be the midpoint of AB. Let M be the intersection of AC and PD and N the intersection of BD and PC. A circle is incribed in the quadrilateral PMSN. Prove that the radius of the circle is MP- MS.
1. Let the diagonals of the square
2. F ind the sum of all the 5-digit integers which are not multiples of 1 1 and whose
digits are 1, 3,
4,
7, 9.
3. Let a1, a2, . . . , an be positive integers, not necessarily distinct but with at least five
si <j s
distinct values. Suppose that for any 1 from
i
and
j such that ai + aj
�
ak +
a.e.
n,
there exist k, £, both different
What is the smallest possible value of
n? 4. A student divides an integer that
m
by a positive integer m ____: n
=
0
·
167ala2
·
·
·
n,
where
n
s 100, and claims
.
Show the student must be wrong. 5. The numbers
f, �, . .. , 20\ 0
are written on a blackboard.
A student chooses any
two of the numbers, say x, y, erases them and then writes down x
+y+
xy.
He
continues to do this until only one number is left on the blackboard. What is this
number?
17
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) (Junior Section, Round
2
2010
solutions)
Let 0 be the centre and r the radius of the circle. Let X, Y be its points of contact with the sides PM , MS, respectively. 1.
Since OY j_ MS and LYSO= LASP=45°, SY=YO=r. Also LOPX = LPDA ( since O P II DA) and LOXP= LPAD= 90°. Therefore 60X P c:::: 6 PAD. Hence OX/XP = PA/AD= 1/2. Hence PX = 2r. Therefore PM - MS = 2r + MX MY-r=r. c
D
p
A
B
First note that an integer is divisible by 11 if and only if the alternating sum of the digits is divisible by 11. In our case, these are the integers where 1,4 and 7 are at the odd positions. Let S be the sum of all the 5-digit integers formed by 1, 3, 4, 7, 9 and let T be the sum of those which are multiples of 11. Then 2.
s=4!(1 +
3 + 4 + 7 + 9)(1 + 10 + 100 + 1000 + 10000)
=6399936
T=2!2!(1 + 4 + 7)(1 + 100 + 10000) + 3!(3 + 9)(10 + 1000)=557568. Thus the sum is 6399936-557568=5842368. are the two smallest values. Then a1 = x and let s be the smallest index such that a8 = y. Now there are two other terms whose sum is x + y. Thus we have a2 = x and as+I = y. Since a1 + a2 = 2x, we must have a3 = a4 = x. Similarly, by considering the largest two values w < z, we have an =an-I =an-2 = an-3 = z and another two terms equal tow. Since there is one other value, there are at least 4 + 2 + 4 + 2 + 1=13 terms. The following 13 numbers 3.
a1 :::; a2 :::;
·
·
·
:::; an. Suppose
x
< y
18
satisfy the required property: 1, 1, 1, 1, 2, value of n is 13.
4. We have
m < 0 167 < 0 168 n ·
·
2, 3, 4, 4, 5, 5, 5, 5.
Thus the smallest possible
167n::; 1000m < 168n.
=>
Multiply by 6, we get
1002n::; 6000m < 1008n
6000m- 1000n < 8n::; 800.
=>
But 6000m -lOOOn 2': 2n > 0. Thus 6000m - 1000n 2': 1000 since it is a multiple of
1000. We thus get a contradiction.
We shall prove by induction that if the original numbers are
5.
the last number is
(1 + a1)
·
·
·
n 2': 2, then
(1 + an) - 1.
The assertion is certainly true for n
=
2, the base case. Now suppose it a1, ..., ak+l written on the board.
is true for
k + 1 numbers After one k numbers. Without loss of generality, we can assume that the student erases ak and ak+l and writes bk ak+ak+l +akak+l (1+ak)(1+ak+l)-l. After a further k operations, we are left with the number n
=
k
a1, ..., an,
2': 2.
Consider
operation, we are left with
=
=
This completes the proof of the inductive step. Thus the last number is
( 1 + �) ( 1 + �) ... ( 1 + 20 10 ) - 1 1
19
=
20 10
Singapore Mathematical Society Singapore Mathematical Oly mpiad (SMO) 2010 (Senior Section)
Tuesday, 1 June 2010
0930 - 1200 hrs
Important: Answer ALL 35 questions. Enter your answers on the answer sheet provided. For the multiple choice questions, enter your answer on the answer sheet by shading the bubble containing the letter (A, B, C, D or E) corresponding to the correct answer. For the other short questions, write your answer on the answer sheet and shade the appropriate bubble below your answer. No steps are needed to justify your answers. Each question carries 1 mark. No calculators are allowed.
PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO
20
Multiple Choice Questions
1.
. F mdthevalue of (A)
(B) (C)
2.
1
(1x2x3)+ (2 x4x6) + (3x6x9)+···+ (335 x670 x1005) (1x3x6)+ (2x 6xl2)+ (3x9x18) +··· + (335 x1005 x20 10)
-
3 2 3 1 -
6
(D)
1 2
(E)
-
4 9
If a, b, c anddare real
numbers suchthat
b+c+d= a+c+d= a +b +d= a +b+c = r, a d b c findthevalue of r. (A)
(B) (C) (D) (E)
3.
If
3 1 -1 3 or 1 3 or-1
1 a a nd tan x+- - =--,where a, b a nd c 2 4 tan x b-;rc are positiv e integers, find thevalue of a+ b +c. O< x<
(A)
(B) (C) . (D) (E)
;r
a nd sin x-cos x=
7r
8
32 34 48 50
21
(A) (B) (C) (D) (E)
5.
104 224 312
336 676
In the figure below, ABC is an isosceles triangle inscribed in a circle with centre 0 and diameter AD, with AB = AC. AD intersects BC atE , and F is the midpoint of OE. Given that BD is parallel to.FC and BC = in em. (A)
(B) (C)
(D) (E)
6.
D A
Find the number of ordered pairs (x, y), where x is an integer and y is a perfect square, such that y = (x-90)2 -4907. (A) (B) (C) (D) (E)
7.
3/5 2 J6 2J3 J7 2./6
2J5 em ,find the length of CD
0
1 2 3
4
LetS= {1, 2, 3, . , 9,10 }. A non-empty subset ofSis considered "Good" if the number of even integers in the subset is more than or equal to the number of odd integers in the same subset. For example, the subsets {4, 8}, {3, 4, 7, 8} and {1, 3, 6, 8,10} are "Good". How many subsets ofSare "Good"? . .
(A) (B) (C) (D) (E)
482 507 575
637 667
22
8.
If the graph of a quadratic function f (x) = ax
2 + bx + c (a i- 0) passes through
two distinct points (r, k) and (s, k), what is j(r + s)? (A)
(B) (C)
(D) (E)
9.
None of the above
Find the number of positive integers k < 100 such that 2(36n ) + k (23n +l) -1 is divisible by 7 for any positive integer n. (A)
10
(B)
12
(C)
13 14 16
(D) (E) 10.
2k c k-c 2k - c
Let
ABCD be
a trapezium with AD parallel to BC and LADC = 90° , as shown in
the figure below. Given that M is the midpoint of AB with CM = BC + CD+ DA
(A)
(B) (C) (D) (E)
=
13
2
em
17 em, fmd the area of the trapezium ABCD in cd.
26 28 30
A
,------,.-,
33 35
B
23
D
c
and
Short Questions
11.
12.
The area of a rectangle remains unchanged when either its length is increased by 6 units and width decreased by 2 units, or its length decreased by 12 units and its width increased by 6 units. If the perimeter of the original rectangle is x units, find the value of x. For r = 1, 2, 3, ..., let
2
1 -+
u,
=1 + 2+3 + ... + r. Find the value of 3 100 + ··· + + . + + + ··· +
(:J (: + J (: : :,) (: : .l
,
13.
If
14.
If
15.
2010!=Mx10k,
,
+
,
,
,
.�.J
.
where Mis an integer not divisible by 10, find the value of k.
1 1 a > b > 1 and -- + -- = -J1229 , find the value of log b 1ogb a a
rl
1 ·
logab b
1
logab a
For any real number X, let X denote the smallest integer that is greater than or
LxJ denote the largest integer that is less than or equal to x (for J1.23l = 2 and LL23J =1). Find the value of
equal to x and example,
I: r 2010 _ 2o10 o
k=ll
x2o1o
l Jl · k
k
.
Fmd the value of + (1_x)2010•
16.
Let f ( x)
17.
If a, b and c are positive real numbers such that
=
x2o1o
ab + a + b = be+b +c ca +c +a = 35, find the value of (a+ 1)(b + 1)(c + 1). =
24
18.
In the figure below, AB and CD are parallel chords of a circle with centre 0 and
radius rem. It is given that AB = 46 em, CD= 18 em and LAOB= 3 x LCOD. Find the value of r. A
B
19.
Find the number of ways that 2010 can be written as a sum of one or more positive integers in non.,.decreasing order such that the difference between the last term and the first term is at most 1.
20.
Find the largest possible value of n such that there exist n consecutive positive integers whose sum is equal to 2010. ·
21.
Determine the number of pairs of positive integers n and m such that 2 1! + 2! + 3! + . + n! = m • .
22.
·
.
The figure below shows a circle with diameter AB. C and D are points on the circle on the same side of AB such that BD bisects LCBA. The chords AC and BD intersect at E. It is given that AE = 169 em and EC = 119 em. If ED = x em, find the value of x.
25
23.
Find the number of ordered pairs (m, n) of positive integers in and n such that m + n = 190 and m and n are relatively prime.
24.
Find the least possible value of j(x) =
25.
Find the number of ways of arranging 13 identical blue balls and 5 identical red balls on a straight line such that between any 2 red balls there is at least 1 blue ball.
26.
LetS= {1, 2, 3, 4, . . . , 100000}. Find the least possible value of k such that any subset A ofSwith IAI = 2010 contains two distinct numbers a and b with Ia- bl::; k.
27.
Find the number of ways of traveling from A to B , as shown in the figure below, if you are only allowed to walk east or north along the grid, and avoiding all the 4 points marked x.
9
1+cos2x over all real numbers for which j(x) is defined.
+
25
1-cos2x
, where x ranges
North
L
East
A
28.
Two circles C1 and C2 of radii 10 em and 8 em respectively are tangent to each other internally at a point A. AD is the diameter of C1 and P and Mare points on C1 and Cz respectively such that PMis tangent to Cz, as shown in the figure below. If PM=
.fiO em and
LPAD
=
xo, find the value of x.
26
29.
30.
31.
Let a, b and cbe integers with a > b > c > 0. If b and c are relatively prime, b + c is a multiple of a, and a+ cis a multiple of b, determine the value of abc. Find the number of subsets {a, b, c} of {1, 2, 3, 4, ..., 20} such that a< b - 1 < c- 3. Let f(n) denote the number ofO's in the decimal representation ofthe positive
integer n. For example-, /(10001 123) = 3 and /(1234567) = 0. Let M = /(1)x2f(I)
+
j(2)x21<2>
+
/(3)x21<3>
+ . . · + f(99999)x21<9999� .
Find the value of M- 100000.
32.
33.
Determine the odd prime number p such that the sum of digits of the number p4 -5 p 2 + 13 is the smallest possible . The figure below shows a trapezium ABCD in which AD II BC and BC = 3AD. F is the midpoint of AB and E lies on BC extended so that BC = 3CE. The line segments EF and CD meet at the point G. It is given that the area of triangle GCE is 15 cm2 and the area of trapezium ABCD is k cm2• Find the value of k.
c
B
34.
Let P (x)= a0 + a1x + a x2 +
E
+
anxn be a polynomial in x where the coefficients a0, a1, a ,•••, an are non-negative integers . If P (1) 25 and P (27) 1771769 , 2 2
· · ·
find the value of ao + 2ar + 3� + ... + (n + 1)an.
27
=
=
35.
Let three circles rl' r2' r3 with centres At, Az, A3 and radii Tt, rz, r3 respectively be mutually tangent to each other externally. Suppose that the tangent to the circumcircle of the triangle A1AzA3 at A3 and the two external common tangents of r1 and r2 meet at a common point P, as shown in the figure below. Given that r1 = 18 em, r2 = 8 e m and r3 = k em," find the value of k.
p
28
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2010 (Senior Section Solutions)
1.
Answer: (A)
(1x2x3) + (2x 4x6) + (3x6x 9) +···+ (335 x 670x 1005) (1x3x 6) + (2x 6x12) + (3x 9x 18) + ···+ (335 x1005 x2010) =
=
2.
(1x2x3)[13 +23 +33 +···+3353] (1x3x6)[13 +23 +33 +···+3353]
�--��--��------�
1x2x3 1 = 1x3x6 3
Answer: (E) From the given equations, we obtain
a + b +c +d = a(r +1), a + b +c +d = b (r +1), a +b +c +d =c(r +1), a +b +c +d =d (r +1). Adding these four equations gives
4(a + b +c +d) = (a + b +c +d)(r +1), that is,
(3-r)(a +b +c +d) =0.
Thus r = 3, or a +b +c +d =0. If a +b +c +d =0, then we see from the original given equations that r = -1 Hence the value of r is either 3 or -1
.
.
3.
Answer: (E)
2
1[ . xcos x= . x-cosx)2 =-, We have (sm which .rmp1"1es that . sm 16
Therefore we obtain
tanx+
1 --
=
tan x
sin x --
cos x
cos x
1
sin x
sin xcos x
+--=
=
32 16-:r2
Hence a+ b+ c = 32+ 16+ 2 = 50 .
4.
Answer: (C) First, we note that 4n3
=
[n(n+ 1)]2 - [n(n- 1)]2 • Thus
29
.
16 -1[2 ----
32
Therefore 143+ 153 + . . . + 243 + 253 142 x152 132 x142 + = 4 4 152 x162 142 x152 + . . .. + 4 4 242 X 252 232 X 242 + 4 4 252 X 262 242 X 2s2 ---
��-
=
4 252 x26 2
4 132 x142
4 = (32 X 13)(18 Thus
5.
= (25 x 13+ 13 x7)(25 x 13-13 x7) 4 X 13) = 9 X 64 X 13 2 •
�143 +153 +163 +...+243 +253
= 3x8x 13 = .312.
Answer: (B) Since the diameter AD perpendicularly bisects the chord BC,
BE = EC =
.JS.
Also, given that BD II FC, we have LDBE = LFCE. Thus MDE is congruent to !1CFE, so DE = FE. As F is the midpoint of OE, we have OF = FE = ED. Let OF = x. Then AE = 5x. Using Intersection Chord Theorem, we have
AExED = BExEC, which leads to 5x2 = 5. Consequently we obtain x = 1. Now CD2 = C£2+ ED2 gives CD =
6.
.J5+1= ../6 .
Answer: (E) Let y = m2 and (x-90)2 = k?-, where m and k are positive integers. Then we obtain k 2 -m2 =4907 =7x701=1x4907, which gives (k -m)(k+m) =7x701 or (k-m)(k+m) =1x4907. It follows that k-m = 7 and k+ m = 701, or k-m = 1 and k+ m = 4907. Solving these two pairs of equations gives (k, m) = (354, 347) and (k, m) = (2454, 2453).
30
Therefore the ordered pairs (x, y) that satisfy the given equation are: (444, 3472 ), (-264, 3472 ), (2544, 24532 ), (-2364, 24532 ). Hence the answer is 4.
7.
Answer: (D) Let the number of even integers in a "Good" subset of S be i, where i = 1, 2, 3, 4, 5, and the number of odd integers in that subset be j, where j 0, 1, 2, ... , i. Then the number of "Good" subsets of S is =
t<(:J�(�) (:}(�) (:) +(�}(�) (:) (�) (�)<(�) (:) G) =
+
+
+
+
+
+. . . +
+. . .
=5(1 + 5) + 10(1 + 5 + 10) + ...+ (1 + 5 + 10 + 10 + 5 + 1) =30 + 160 + 260 + 155 + 32 =637.
8.
Answer: (B) Let g(x) = f (x) -k. Then g(r)
f (r)- k = k-k = 0. Similarly, g(s) = 0. Therefore r ands are roots of the quadratic equation g(x) =ax2 +hx+c-k =0, from which =
b
we deduce that r + s = - -. Hence a
b a
b a
b a
j(r + s ) = j(--) = a (-- )2 +b(--) +c = c.
9.
Answer: (D) We have 2(36n) +k (23n+l) -1= 2(272n) +2k(8n) -1 = 2( -1)2n +2k(ln) -1 (mod 7) = 2k +1 (mod 7). l Thus, for any positive integer n, 2(36n) + k(23n+ ) -1 is divisible by 7 if and only
if 2k +1 = 0 (mod 7 ). As k< 100, it is clear that the congruence holds for k =3, 10, 17, . .., 94. Thus the required number of positive integers k is 14. 10.
Answer: (C)
31
'
i
Extend DA and CM to meet atE as shown in the figure above. Since AM = MB, LAEM = LBCM and LAME = LBMC, we conclude that �M is congruent to tillCM. Therefore AE = BC and CM=EM. Thus CE = 2CM= 13.
Let the area of trapezium ABCD be S cm2 . Then S =_!_ (DE)(DC), and we have
2 (DE + DC)2 =DE2 + DC2 + 2(DE)(DC) =C£_2 + 4S =132 + 4S =169 + 4S. Now DE+DC=DA+AE+DC=DA+BC+DC=17. Hence 172 =169 + 4S, and it follows that S = 30. 11.
Answer: 132 Let the length and width of the original rectangle beL and W respectively. Then
LW= (L + 6)(W -2)
and
LW = (L-12)(W + 6).
Simplifying the above equations, we obtain
L-2W=12
and
3W-L=6 .
Solving the simultaneous equations, we get L=48 and W=18. Hence the perimeter of the rectangle is 132 units.
12.
Answer: As ur
2575
r r + 1) , we have =1 + 2 + 3 + ... + r = ( 2 2 =t �-� =2-__2_=�. =t t_!_ +1 +1 r�l r ( r + 1) r�l r r + 1 r�l ur
(
) i i i:-/ =i:( : J=i:(i +12 )=_!_ ( (n2 1) + n)=�(4 n + 3).
Hence, Sn : =
i i�l i+l
i�l "" __L i�l L...J u, r�l In particular, slOO =2575.
13.
2
n
+
Answer: 501 The number k is the number of the factor 10 that occurs in 2010!. This number is given by the number of pairs of prime factors 2 and 5 in 2010!. Now between 1 and 2010, there are: • • • •
402 integers with 5 as a factor; 80 integers with 25 as a factor; 16 integers with 125 as a factor; 3 integers with 625 as a factor.
32
Therefore the total number of prime factor 5 in 2010! is 402 + 80 + 16 + 3 =501. As there are clearly more than 501 prime factor 2 in 2010! , we obtain k =501.
14.
Answer: 35 First note that since a> b > 1,
1
--
logabb
-
1 logab a
1
1
logabb
logab a
·
>
0. Then
= Iogb ab -loga ab = (Iogb a+ 1) (loga b +1) = logb a-loga b
1
1
loga b
Iogb a
·
=-----
[�-�)' = [�+ : J { : X : J =
-
lo , a
lo . b
lo , a
= ..)1229-4
= .J1225
= 35.
15.
Answer: 1994
l� J
If
= 2010 - 2 10 =0, so Xl= .If k 12010, then X: r O k 2 10 - 2 10 y k ' 2010, then 0 < y:= < 1 so r l =1.
Consider k
=
1, 2, ..., 2010.
� l� J
'
Since the prime factorization of 2010 is 2 x 3 x 5 x 67, we see that 2010 has 16 distinct divisors . Hence
o l2010
Ik=l I 16.
k
l Jl = klL20!0JXl+kl2:20!01 l y
- 2010 k
=number of non-divisor of 2010 among k =2010 -16 =1994.
Answer: 1005 _
Observe that j(x) + j(l -x)-
It follows that
(l- xyolO x2o1o -1 . + x2oJo+ (1-x )2010 (1-x )2010+x20JO _
33
=
17.
1005.
Answer: 216 Adding 1 to both sides of the given equation ab + a + b
=
35, we obtain
(a+ 1)(b+ 1) = 36. Likewise, adding 1 to the other two given equations gives (b+ 1)(c+ 1) = 36 and (c+ 1)(a+ 1) = 36. Now multiplying the three resulting equations above leads to 3 [(a+1)(b+1)(c+1)] 2 = 36 = 66• 3 It follows that (a+1)(b+1)(c+1)=6 =216.
18.
Answer:
27
A
B
Let M and N be the midpoints of AB and CD respectively and let L CON = x. Then LAON=3x and 3 23 AM r s 3x 3sin x � 4sin x = =3_4sin 2 x . = = sm x r sm x 9 CN 23 1 1 1 . Thus sm 2 x = "4 3 g = g , and so sm x= "3 . Hence r=
--
(- )
�
.
CN =27. sin x
34
19.
Answer: 2010 Consider any integer k where 1 :::; k:::; 2010. By division algorithm, there exists unique pair of integers (q, r) such that 2010 = kq+ r with 0:::; r:::; k- 1. We rewrite this as 2010 = (k- r)q+ r(q+ 1). That is, k- r copies of q and r copies of q+ 1 add up to 2010. Thus there is one desired expression for each value of k, which is clearly unique . Hence there are 2010 such expressions in all .
20.
Answer: 60 Let a be a positive integer such that the sum of n consecutive integers a, a+ 1, ... , a+ (n- 1) is 2010, that is,
a+(a+1)+···+(a+n -1) 2010. =
n(2a+n -1) . =2010, or Th.I S gives
2 n(2a+n -1) =4020=22 x3x 5x67.
(1)
Since n < 2a+ n- 1, we have
n<.J22x3xSx67 =2.JlOOS <2x 32=64.
(2)
Now (1) and (2) imply that n E{1, 2,3,4,S,6,10,12,1S,20,30,60}. Sincenand 2a+ n- 1 have different parities, it follows that nand we have If n
21.
=
4020 n
have differ�nt parities. Consequently,
n E{1,3,4,S,12,1S,20,60}.
60, then 2a+n -1 =
4020 =67, so a 4. Thus the largest possible value of n is 60. 60 =
Answer: 2 First note that if n 2: 4, then
1!+2!+3!+···+n! =1!+2!+3!+4! (modS) =1+2+1+4=3 (modS).
Since the square of any integer is congruent to either 1 or 4 modulo S, it follows that 1 ! + 2! + 3! + + n! i= m2 for any integer m in this case. So we consider n< 4. Now we have · · ·
1!
=
12,
1! + 2!
=
3, 1! + 2! + 3!
=
32•
Hence we conclude that there are two pairs of positive integers (n, m ) , namely, (1, 1) and (3, 3 ), such that 1! + 2! + 3! + + n! = m2• · · ·
35
22.
Answer: 65 EC 119 = = . Thus we can let BC = 119y BA EA 169 and BA = 169y for some real number y. Since LBCA = 90°, we have Since BE bisects LCBA, we have
BC
AB2 = AC2+ B C2
(169yy =(169+119) 2 +(119y) 2 y 2 (169 -119X169+119)= (169+119) 2 y2 =
169+119
169-119 12 y=5
=
144 25
Hence, from triangle BCE, we have BE =
�119 2 + (119y) 2 =119 x 13 . 5
Finally, note that f).A[)E and MCE are similar, so we have AE xCE 169x119 =65 cm. = ED= 119x 153 BE
23.
Answer: 72 First we find the number of ordered pairs (m, n) of positive integers m and n such that m+ n = 190 and m and n are not relatively prime. To this end, write m = ka and n = kb, where k, a and b are positive integers with k > 1. Since m+ n = 190, we see that k is a factor of 190 = 2x5x19 with k t 190. We consider six cases: (i) (ii) (iii) (iv) (v) (vi)
k = 2. Then a+ b = 95, and there are 94 such pairs (a, b) of a and b such that the equation holds. k = 5. Then a + b = 38, and there are 37 such pairs (a, b) of a and b such that the equation holds. k = 19. Then a + b = 10, and there are 9 such pairs (a, b) of a and b such that the equation holds. k = 10. Then a+ b = 19, and there are 18 such pairs (a, b) of a and b such that the equation holds. k = 38. Then a+ b = 5, and there are 4 such pairs (a, b) of a and b such that the equation holds. k = 95. Then a+ b = 2, and there is 1 such pair (a, b) of a and b such that the equation holds.
(
It follows from the above cases that the number of ordered pairs m, n) of positive integers m and n such that m+ n = 190 and m and n are not relatively prime is 94+37+9-18-4-1 = 117.
36
Since the total number of ordered pairs (m, n) such that m + n = 190 is 189, we conclude that the required number of ordered pairs (m, n) where m and n are relatively prime is 189 - 117 = 72.
24.
Answer: 32 For all real values of x for which] (x) is defined, we have
9
j(x) =
l+cos 2x 1-cos 2x
= _!_
2
= 17 + _!_
� t an2
� 17 +
2
=
9
=
� t an2 x+9) + _!_ (25 cot
= 17
+
32.
2
� ( ��
2
X
+ 25 coe
2
X
x+ 25
)
) ))
X
t an2 x 25cot2 x
(AM-GM Inequality)
least possible value of j(x) is 32. fi) = 32. Thus the .
�3
Answer: 2002 First we place the 5 red balls on a straight line and then place 1 blue ball between 2 adjacent red balls. With this arr angement fixed, the condition of the question is satisfied. We are now left with 9 blue balls. We can place the remaining 9 blue balls into the spaces before, after or in between the 5 red balls. The number of ways that this can be done is the answer to the question. Including the two ends, there are 4 + 2 = 6 spaces into which these 9 blue balls can be placed. The number of ways of distributing the 9 blue balls into the 6 spaces is 4
( : J ( � J c :J 9
26.
25
+ 2cos2 x 2sin2 x
_!_(2J9x 25) 2
Note that j(t an-1
25.
25
+
+
-!
�
�
�
2002.
Answer: 49 Consider the following subsets of S: S1
= {1, 2, 3, ..., 49, 50},
{51, 52, 53, ..., 99, 100}, s3 = {1 01, 102, 1 03, ..., 149, 150}, S2 =
Szooo =
{99951, 99952, 99953, ..., 99999, 100000}.
In other words, Si = {50i- 49, 50i- 48, ..., 50i} fori= 1, 2, ..., 2000. Note that Sis partitioned into these subsets St, Sz, S 3 , ... , Szooo.
37
By Pigeonhole Principle, for any subset A of S with where 1:::; i:::; 2000, such that I AnSi � I 2. Let
a,b EAnsi. It
is clear that
IAI = 2010,
there exists i,
Ia- bl:::; 49.
To show that 49 is the least possible value of k, we find a subset A � S with lAI = that Ia- bl2: 49 for any distinct a,b EA. Let
2010 such
A={ 49j+1: j=0,1, 2, ... , 2009}={1,50, 99,1 48, ... , 98442}.
Then A is a subset of S with
27.
Answer:
IAI = 2010
and
Ia- bl2: 49
for any distinct
a,b EA.
11 2
E �--+---+---4----r--� n
Figure 1 A
c
We observe that to avoid the four points marked x, the path must cross either C, D orE as shown in Figure 1 above. Further, the paths that cross C, D orE are exclusive, that is, no path can cross both C and D or D and E, or C and E. There is only 1 way to get from A to C and from A toE. It is easy to see that there are 4 ways to get from A to D. 1
1 1
1
1
c
6
5
..
4 .. 3
2 1
21
• • •
B
15
10 6
Figure 2
3
1
To count the number of ways to get from either C, D orE to B, we note that the number of ways to get to a certain junction is the sum of the numbers of ways to get to the two junctions immediately preceding it from the left or from below (as shown in Figure 2). Therefore there are 21 ways to get from C to B. Similarly, there are 21 ways to get from D to Band 7 ways to get from E to B. Hence the number of ways to get from A to B that • pass through C: number of ways from A to C x number of ways from C to B= 1 X 21 = 21; • pass through D: number of ways from A to Dx number of ways from D to B= 4 X 21 = 84; • pass through E: number of ways from A to E x number of ways from E to B=1x7=7. It follows that the total number of ways from A to B is 21 + 84+7= 1 12. 38
28 .
Answer: 60
Let 0 be the centre of
C2
and let PA intersect
C2 atB .The
homothety centred atA
It mapsB toP. Thus AB = !__ (This 10 AP 10 can also be seen by connecting P to the centre 0' of C1 so that the triangles ABO and APO' are similar.) The power of P with respect to C2 is PM2 = 20. Thus mapping
Cz
to
C1
has similitude ratio
!__
PB·PA = 20,or equivalently (PA - AB)PA =20. Together with AB = !_ ,we AP 10 obtain AB =8 and AP = 10 . Consequently,the triangle ABO is equilateral,and hence LPAD = LBA0 = 60°.
29.
Answer: 6 We shall show that a =3,b =2 and c =1. Note that 2a > b + c.As b + c is a multiple of a ,it follows that a =b + c. Let a + c =kb . Then kb =a + c = b + c + c,so 2c = (k- 1)b . Since c < b,we must have k =2 and therefore b =2c. Since b and c are relatively prime,this implies that c = 1 and b =2 . Thus a =3 . Hence abc = 6.
30 .
Answer: 680 For any 3-element subset {a ,b,c},define a mapping f by f ( {a ,b,c}) = {a ,b - 1,c - 3}. Now observe that {a ,b,c} is a subset of {1, 2, 3, 4, . . . , 20} with a < b - 1< c - 3 if and only if j( {a ,b, c} ) is a 3-element subset of { 1, 2, 3, . . . , 17}. Hence the answer is
C:)
= 680.
39
31.
Answer: 2780 Note that 0 ::;; f(n)::;; 4 for 1 ::::;n::::; 99999. Fork= 0, 1, 2, 3, 4, let ak denote the number of integersn, where 1 ::::;n::::; 99999, such thatj (n) = k. Then 4 4 k k M= kak 2 = kak 2 . k =O k =l.
L:
L:
By considering the number of 2-digit, 3-digit, 4-digit and 5-digit positive integers with exactly one 0 in their decimal representation, we obtain � = 9+ 9 X 9 X 2+ 9 X 9 X 9 X 3+ 9 X 9 X 9 X 9 X 4=28602 .
Similarly, we have
a =9 + 9 X 9 X 2
( �)
+9 X 9 X 9 X
(�)
= 4626,
a3=9+9 2 x4=333, a =9. 4 Hence 3
M =1x28602x2+2x4626x22 +3x333x2 +4x9x24 =102780,
and it follows that M- 100000 = 2780. 32.
Answer: 5 Let n = p4 -5 p 2 +13. When p = 3, we haven= 49 and so the sum of digits is 13. When p= 5, we haven= 513 and the resulting sum of digits is 9. Now let p > 5 be a prime. We have n= p4 -5p 2 +13= ( p-2)( p-1)( p+1)( p+2)+9. Since p- 2, p- 1, p, p + 1 and p + 2 are five consecutive integers, at least one of them is divisible by 5. Since pi= 5, we must have 5 divides one of the numbers p- 2, p- 1, p + 1 and p + 2, so 5 divides the product (p- 2)(p- 1)(p + 1)(p + 2). Observe that at least one of the numbers p + 1 and p + 2 is even. Therefore we see that (p- 2)(p- 1)(p + 1)(p + 2) is divisible by 10. It follows that for any prime p > 5, the number n has at least two digits and the units digit is 9, so the sum of the digits is greater than 9. Consequently the smallest possible sum of digits is 9, and this value is attained when p= 5.
40
33.
Answer: 360 X
ExtendBA andCD to meet at X .LetH be the point onCD such thatFH II Let AD
=
CE
=
a. Then
BC
=
ECG,
3a , andFH
( ) CE
=
1
2
(AD +
BC)
By the similarity of trianglesFHG and we have 2 FH x area of (i) area of FHG = (ii)
Ll CG CE
HG
FH
Ll ECG
=
It follows from (i) and (ii) that the area of triangleFDH
(iii)
XA
XF
AD
XAD LlXAD ( )
60 cm2;
=
3
2
� x 60
2
HG
=
.
90 cm2 .
bey cm2 . By the similarity of triangles XAD and
1
- =- =
FH 2 area of ilXDF
(iv) area of LlXFH
, so that XA =AF and hence =
2x area of AD
FH
=
2
y
=
=
2y cm2;
4y.
It follows from (iii) and (iv) that the area of triangleFDH 2 Since the area of triangleFDH is 90 cm , we get y = 45. Finally, by the similarity of triangles andABC, area of
2a.
2 , so thatHG = 2CG andDH =HC =
-=-=
Now, let area of triangle XFH , we have
=
BC.
LlXBC (BC)2 =
AD
Hence the area of trapeziumABCD
XAD
y =9y . =
8y = 360 cm2•
41
=
4y-2y
=
2y cm2 .
34.
Answer: 75 First we note that every positive integer m can be written uniquely (in base 27) as m=b0 +b1 x27+b2x27 2 +···+b7 x277, where r and b0, bp b2,
• • •
, br (depending on m) are non-negative integers with
bi < 27 for i =
0, 1, 2, ..., r. Since the coefficients of P(x) are non-negative integers and P(1) = 25, we see that ak ::;; 25 < 27 for 0:::; k:::; n. Thus by the above remark, the polynomial P (x) is
uniquely determined by the value P(27) = 1771769. Writing 1771769 in base 27, 3 we obtain 1771769 = 2+11x27+9x27 +3x274• Therefore 3 P (x) =2+11x+9x +3x4• Hence a0 +2a1 +3a +···+(n+1)an =2+2x11+4x9+5x 3=75. 2 35.
Answer: 12
p
N,. Let P be the point of concurrence of the
First we shall show that r3 =
tangent to the circumcircle of the triangle ��As at As and the two external common tangents of f1 and f • Note that the line joining � and � also passes 2 through P. First we have PA; =P� · PA1 , so
( ) PAs
p�
2
=
PA1
PAs =..2. That is, = p� r2 p�
On the other hand, PAs sinLAg� P sin� r1 +r3 = -= =-sin�As P sin� r +r3 p� 2 1+ 3 Thus = ' ' • Solving for r3, we obtain r3 = + 2 ' ,3 r2 =8 , we get r3=12.
--
N,.
3 v;;
42
3. v;;
Substituting r1 =18 ,
',
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2010 (
. Senior Section, Round 2) Saturday, 25 June 2010
0930-1230
INSTRUCTIONS TO CONTESTANTS
1.
2.
3.
Answer ALL 5 questions. Show all the steps in your working. Each question carries
10
mark.
4. No calculators are allowed. 1. In the triangle ABC with AC > AB,
D is the foot of the perpendicular from A onto BC and E is the foot of the perpendicular from D onto AC. Let F be the point on the line DE such that EF DC = BD DE Prove that AF is perpendicular ·
·
.
to BF. 2. The numbers
t, �, . . . , 2 0\ 0
are written on a blackboard.
A student chooses any
two of the numbers, say x, y, erases them and then writes down x
+
y
+
xy. He
continues to do this until only one number is left on the blackboard. What is this number? 3. Given
a1 2': 1 and ak+l ;::: ak + 1 for all k 1, 2, , n, show that af + a� + + a � 2': (a1 + a2 + + an) 2 . ·
·
·
·
·
·
4. An infinite sequence of integers, a0 , = for any n 2': 0, where
an, a1
·
=
an+l an - bn,
·
·
a1, a3 , with a0 > 0, has the property that bn is the number having the same sign as .
.
•
but having the digits written in the reverse order. For example if =
1089 and
all n;:::l.
a2
=
-8712, etc. Find the smallest value of
a0
5. Let p be a prime number and let
so that
a0
=
1210,
an i= 0 for
a1, a2 , ... , ak be distinct integers chosen from n 1, 2, ..., p 1. For 1 ::; i ::; k, let r } ) denote the remainder of the integer nai upon n division by p, so 0 ::; r} ) < p. Define n S { n : 1 :S n :S p - 1, rin) < < r )} -
·
=
Show that
S has less than k2_f1
elements.
43
·
·
k
.
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) (Senior Section, Round
2
2010
solutions)
Since we are supposed to prove LAFB=goo, it means that the 4 points A, B, D, F are concyclic. Note that AC > AB implies that LB > LC. If TD is the tangent to the circumcircle w of the triangle ABD with B and T lying opposite sides of the line AD, then LADT = LB > LC = LADE so that w intersects the interior of DE at F. Therefore F can only be in the interior of DE. Now observe that the triangles ADE and DCE are similar so that AD/AE =DC/DE. By the given condition, this can be written as AD/AE = BD/ EF. This means the triangles ABD and AFE are similar. Thus LABD = LAFE. This shows that A, B, D, F are concyclic. Therefore LAFB=LADE=goo. 1.
A
c
B
2.
See Junior Section Question 5.
We will prove it by induction. First, it is clear that suppose it is true for n terms. Then 3.
ar 2:: af since a1 2:: 1.
n+l n n La� 2:: a�+l +La� � a�+l + (Lak) k=l k=l k=l n n+l (Lak) +a�+1-a�+1-2an+1Lak. k=l k=l 2
2
=
44
Next,
a�+l-a;+l-2an+l.L.:�=lak :?: 0. Since ak+l- ak :?: 1, we have a�+l- a� :?: ak+l + ak. Summing up over k 1, ... , n, and using ar - al :?: 0, we have n a�+l- at:?: an+l+ 2 L ak - a1 k=l To complete the induction, we'll now show that
=
4. If
ao has a single digit, then a1 0. Thus a0 has at least 2 digits. If a0 ab 10a+b, then a 1 9(a-b) which is divisible by 9. it follows that all subsequent terms are divisible by 9. Checking all 2-digit multiples of 9 shows that eventually 9 appears ( Note that ab and ba give rise to the same sequence, but with opposite signs ) : 81 63 27 45 9. =
=
=
=
---+
---+
---+
---+
ao abc, then a1 99(a - c). Thus if suffices to investigate 3-digit multiples of 99, 198, ..., 990. Here we find that 99 will eventually appear: 990 891 693 297---+ -495 ---+ 99. If ao =abed, then a1 999(a-d)+90(b-c). If b, care both 0, then a1 and all subsequent terms are multiples of 999. However, if such numbers appear in the sequence, eventually 999 will appear: 9990 8991 6993 2997 -4995 999. For 1010, we get 909 and for 1011 we get -90. For 1012, we get 1012 -1089 -8712 6534 2178 -6534 and the sequence becomes periodic thereafter. Thus the smallest a0 1012. If
=
=
i. e. ,
---+
---+
---+
=
---+
---+
---+
---+
---+
---+
---+
---+
---+
---+
=
5. Let
rbn) 0 and Tk�1 =
S' Note that
=
=
p.
Set
k {n: 1:::; n:::; p - 1, .2.: lr���- r�n)l p}. i=O =
k
L lr���- r�n) I i=O
=
P
iff
Tbn)
:S;
rin)
:S;
·
·
·
:S;
rk�1.
n E S', l r��)1 - r�n) I r���- dn) n(ai+l- ai) (mod p) and p f (ai+l- ai), the numbers r��)1- r�n), 1 :::; n:::; p- 1, are all distinct. Therefore Thus
lSI
Pl8'1
=
=
IS' I ·
Since for
=
lri�i- ri"ll t I.; lri�i- rl"ll t�j t nEB' i=O i=O nEB' i=O j=l I.;
Therefore
lSI
�
<
�
k2_J1.
45
�
�
(k + l)IS (IS'I + l).
Singapore Mathematical Society
( SMO ) 2010 1)
Singapore Mathematical Olympiad
( Open Section,
Round
0930-1200 hrs
Wednesday, 2 June 2010
Instructions to contestants 1. Answer ALL 25 questions. 2. Write your answers in the answer sheet provided and shade the appropriate bubbles below your answers.
3. No steps are needed to justify your answers. 4. Each question carries 1 mark.
5. No calculators are allowed.
PLEASE DO NOT TURN OVER UNTIL YOU ARE TOLD TO DO SO
46
8n3- 96n2 +360n- 400 . is an integer . Find 1. Let S be the set of all mtegers n such that 2n - 7 the value of lnl.
L
nES
2. Deter mine the largest value of x for which
�
lx2 - 4x - 396011 lx2 +4x - 3960 11.
3. Given that find the value of L
�
X=
1 1 1 L1 /3j + L2 /3j + L3 /3j
+ ... + L7999
1 /3j,
J, where LYJ denotes the greatest integer less than or equal to 1 0 (For example, L2.1J = 2, L30J = 30, L- 10.5J = - 1 1.)
4. Determine the smallest positive integer C such that 6n.�
:::;
y.
C for all positive· integers n.
.
5. Let CD be a chord of a circle r1 and AB a diameter of r1 per pendicular to CD at N with AN> NB. A circle r2 centr ed at C with radius CN intersects r1 at points P and Q, and the segments PQ and CD intersect at M . Given that the radii of r1 and r2 ar e 6 1 and 60 r espectively, find the length of AM. 6.
50
Deter mine the minimum value of
L Xk, k=l
positive number s x1, . . . , X5o satisfying
where the summation is done over all possible
50
L __!_ = 1. X k=l
k
4
7. Find the sum of all positive integer s p such that the expression (x- p)(x- 13 ) + can be expressed in the for m ( +q) ( +r ) for distinct integers q and r.
x
x
� - :2 - : 2
8. Let Pk = 1 + , wherek is a positive integer . Find the least positive integer n 3 such that the product P P Pn exceeds 20 10. 3 ·
·
·
9. Let B be a point on the circle centred at 0 with diameter AC and let D and E be the circumcentres of the tr iangles OAB and OBC respectively. Given that sin L.BOC = g and AC = 24, find the area of the tr iangle BDE. 10. Let f be a real-valued function with the rule f(x) = x3 +3x +6x+1 defined for all real value of It is given that a and bare two r eal numbers such that f(a ) = 1 and f(b) = 19. Find the value of ( a +b
x.
1 1. If cot a +cot ,6 +cot'/'
2
4
)2.
=
- 5,4 tan
cot a cot ,6 +cot ,6 cot')' +cot')' cot a
=
=
17 and B 1 -5, find the value of tan(a +,6 +')').
a +tan ,6 + tan')'
7
47
12.
The figure below shows a road map connecting two shopping malls A and B in a certain city. Each side of the smallest square in the figure represents a road of distance 1km. Regions C and D represent two large residential estates in the town. Find the number of shortest routes to travel from A to B along the roads shown in the figure. 8 km
.B
c 10 km
D A
1 3.
Let
a1
=
+ an> 2
14.
1, a2
=
2
and for all
·���� for all
n 2m,
n 2 2, an+l
where
m is
=
n-1 2n an + + 1 an-1· n n 1
It is known that
a positive integer. Find the least value of m.
It is known that
V9 - 8sin 50° =a + b sinc0 for exactly one set of positive integers
(a, b,c
+ ), where 0 < c < 90. Find the value of b c.
15. If a is a real root of the equation x5 -x3 + x -2 the least positive integer not exceeding x.
a
=
0, find the value of
La6 J, where Lx J is
16. If a positive integer cannot be written as the difference of two square numbers, then the integer is called a "cute" integer. For example, 1, 2 and 4 are the first three "cute" integers. Find the 2010th "cute" integer.
( Note: A
the square of a positive integer. As an illustration, 16 are the first four square numbers.) square number is
1, 4,
9 and
17. Let f (x ) be a polynomial in x of degree 5. When f (x ) is divided by x - 1, x- 2, x - 3, x -4 and x2 -x -1, f (x ) leaves a remainder of 3, 1, 7, 36 and x- 1 respectively. Find the square of the remainder when f (x ) is divided by x + 1.
48
18. Determine the number of or dered pairs of positive integer s (a, b) satisfying the equation 100( a +b)= ab- 100.
( Note:
As an illustr ation, ( 1, 2) and ( 2, 1) are considered as two distinct ordered pairs. )
19. Let p= ab +ba If a, b and p are all prime, what is the value of p? . 20. Determine the value of the following expression:
l J l
J l
J l
l
J
J
11 X 3 11 11 X 2 11 X 4 11 X 2009 +. . . + + + + 2010 2010 2010 2010 2010 '
where
LYJ
denotes the gr eatest integer less than or equal to y.
21. Numbers 1, 2, . . . , 2010 are placed on the circumference of a cir cle in some order. The numbers i and j, where i =F j and i,j E { 1, 2, . . . , 2010} form a friendly pair if
( i) i and j are not neighbour s to each other, and ( ii) on one or both of the arcs connecting i and j
along the circle, all numbers in between
them are greater than both i and j.
Determine the minimal number of friendly pairs. 2 2. LetS be the set of all non-zer o r eal-valued functions f defined on the set of all r eal numbers such that f(x2 +yf(z)) = xf ( x ) +zf(y) for all r eal number s x, y and z Find the maximum value of£( 1 2345) , where f E S. . 23. All possible 6-digit number s, in each of which the digits occur in non-increasing order from left to r ight ( e. g. , 966541) , are written as a sequence in increasing order ( the first thr ee 6-digit number s in this sequence are 100000, 110000, 111000 and so on) . If the 2010th number in this sequence is denoted by p, find the value of L J, where LxJ denotes the greatest integer less than or equal to x.
:0
24. Find the number of permutations a1 a2a3a4a5a6 of the six integers from 1 to 6 such that for all i from 1 to 5, ai+l does not exceed ai by 1. 25. Let A=
( ) c�� )) ( ) C )) ( ( ) c )) ( ( ( (����) - G���)) ( ) () ( (;) ) 2010 0
o
2
+
O10 o
2010 1
2
+
010 1
20 10 2
2
2
+. . . + Determine the minimum integer s such that sA �
( Note:
For a given positive integer n,
other values of r, define
n r
=
= 0.
49
40 20 2010
·
n! _ r ).' for r = 0, 1, 2, 3,
' r. n
·
·
·
,
n; and for all
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) (Open Section, Round
1)
2010
1. Answer: 50 27 8n3- 96n2 + 360n- 400 . 4n2- 34n + 61 + _ . Smce Note that 4n2 - 34n + 61 is 2n _7 2n 7 an integer for all integers n, we must have that 27 divisible by 2n- 7. Hence, 2n - 7 -1,1,-3,3,-9,9,-27,27, so that n 3,4,2,5,-1,8,-10,17. Hence the required sum equals 50. =
=
=
2. Answer: 199 By direct computation,
lx2- 4x- 396011 � lx2 + 4x- 396011 -¢:::=:?- (x2- 4x- 39601)2- (x2 + 4x- 39601)2 � 0 -¢:::=:?- 2(x2- 39601)(-8x) � 0 -¢:::=:?- x(x + 199)(x - 199):::; 0, we conclude that the largest possible value of
x is 199.
3. Answer: 1159 Note that X
-
L11/3j + L21/3j + L31/3j + ... + L79991/3j L Lkl/3j + L Lkl/3j + L Lkl/3j + ... +
L:
L:
1+
2+
L:
3 + ... +
L:
L
Lkl/3j
19
(23- 13) + 2 (33- 23) + 3 (43- 33) + ... + 18 (193- 183) + 19 (203- 193) 19 19(8ooo) - L: k3 k=l 19 20 2 19(8000) -
(
X
:. LlOOJ
)
;
115900 =
1159.
50
4. Answer: 65 6n Define f( n) =I for n=1, 2, 3, ···. It is clear that f( l ) =6, f( 2) =18, f(3) =36, f( 4 ) =54 n.
and f(5) =64.8. For all. n ;:::: 6, f( n) =
6n n!
�
6 6 6 6 6 � l X 2 X J X Ll X S X 6 =6 X 3 X 2 X 1.8=64.8
5. Answer: 78 Extend DC meeting r2 at H. Note that DN =NC =CH = 60. Since M is of equal power with respect to r1 and r2 Thus MN ·MH = MC ·MD . That is MN(MC+60) = . MC(MN +60) giving MN =MG. Thus MN =30. A
The power of N with respect to r 1 is DN · NC = 602, and is also equal to N A ·N B = NA·(AB -NA) =NA·(1 2 2 -NA). Thus NA·(1 2 2 -NA) =602. Solving this quadratic equation, we get NA = 7 2 or 50. Since NA > NB, we have NA = 7 2. Consequently AM =vfNA2+MN 2=vf7 22+302=78. 6. Answer: 2500 By Cauchy-Schwarz inequality,
and equality holds if and only if Xk =50 fork=1, . . . , 50. Therefore the required value is 2500. 7. Answer: 26 Let (x- p)(x- 13)+4=(x+q)(x+r ) . Substituting x=-q into the above identity yields (-q - p)(-q - 13) =-4, which becomes (q+p)(q+ 13) =-4. Since p and q are integers, we must have the following cases:
( a) ( b)
q+p=-4, q+13=1;
( d)
q+p=- 2, q+13= 2
q+p=4, q+13=-1;
(c) q+p= 2,q+13=- 2;or
51
() q = -14,p= 8 and hence (x- p)(x- 13)+4 = (x - 14)(x- 17); (b ), we obtain q = -12,p = 8 and hence (x- p)(x- 13)+4 = (x- 9)(x- 12); (c ), we obtain q =-15,p= 17 and hence (x- p)(x- 13) +4= (x- 15)2; which
For case a , we obtain -For case
For case is NOT what we want;
()
For case d , we obtain q = - l l,p = 9 and hence (x- p)(x- 13)+4 = (x- 11)2; which is also NOT what we want. Hence the two possible values of pare 8 and 18, the sum of which is 26.
8. Answer: 8038 First, note that
( ) (1+.!.k ) 2= (k - 1)(k+1)2 k3
! ! ]__ ]__ Pk = 1+ k _ k 2 _ k3= 1 - k Therefore,
1· 32 2· 42 3· 52 (n- 1)(n+1)2 (n+1) 2 = P 2P3··· Pn=� ·�·43··· n3 4n (n+1)2
> 2010 is equivalent to n2-8038n+1 > 0, which is equivalent Next, observe that 4n to n(n- 8038) > -1. Since n is a positive integer, the last inequality is satisfied if and only if n 2:: 8038. Consequently, the least n required is 8038.
9. Answer: 45 Let d = AC = 24. First, it is not difficult to see that LDEB =LACE and LEDB = LCAB, so that the triangles DEE and ABC are similar.
Let M and N be the feet of the perpendiculars from B onto DE and AC respectively. As M is the midpoint of OB, we have BM = �· Also BN = BO sinLBOC = � x � = 2r Therefore DE = AC x �lfr = d x �� = 5 . Thus the area of the triangle BDE is ! x BM x DE=! x � x 58d = s;;. Substituting d = 24, the area of the triangle BDE is
l
45.
52
10. Answer : 4
We note that f(x) = (x + 1)3 +3(x+ 1 ) + 10. Let g(y) = y3 + 3y, which is a str ictly increasing odd function. Hence f(a) = 1 implies g(a + 1) =-9 and f(b ) = 19 implies g(b +1) =9. Thus, a+1=-(b + 1), implying a+b =- 2, so that (a+b)2 =4.
11. Answer : 11 Let x=tan a, y =tan/3 and z=tan'Y. Then xy+yz+zx xyz x+y+z x+y+z xyz
-
-
4 5 17
6
-
-
17 5
-� and xy+yz +zx=�· It follows that
From the above three equations, xyz= tan(a +pa +'Y) =
x +y +z- xyz 17/6- (-5/6) = =11. 1- 2/3 1- (xy+yz+zx)
1 2. Answer: 2 20 23
Include the points E, F, G, H, I, J, K, L, M, Nand Pin the diagr am as shown and consider all possible routes: E
8km F
G H
10 km
B
c J
I
L
K
M
D A
For the route A
N
p
---+ E ---+ B,
there is 1 way.
For the route A ---+
F ---+
B, there are
For the route A ---+
G---+
B, there ar e
C2) · (D c ) (�) ° 2
53
·
=80 ways. =1260 ways.
For the route
A� H �I� B,
For the route
A� J �I� B,
For the route
A� J
�
K � B,
A� L �I� B,
For the route
A� L � K � B, A� L � M
�
=
=
·
=
=
there are
·
=
·
For the route
A� P � B,
there are
=
=
ways.
ways.
ways.
11 ways.
22023 ways.
Hence, by adding up, there are altogether
4021 an+l - an=
Rearranging the recurrence relation yields
n � 3, we have an- an-1 =
ways.
·
B, there are
there are
ways.
1176 ways.
there are
A� N � B,
for
=
there are
For the route
1 3. Answer:
ways.
there are
For the route
For the route
(�) · (�) 980 (�) · (D · (�) 5880 (�) (D (!) · (�) · (�) 7350 (!) (D (D 4410 (!) (�) 490 (�) · c21) 385 ell)
there are
-ll
n (an- an-1) for n � 2. Thus, n+ .
. n- 2 n- 2 n- 3 (an-1- an-2) = an-2- an-3 ) ( · n n n- 1. n- 2 n- 3 2 1 a1) ··· = n · n -1 ··· 4 · 3( a22 2 2 -n(n- 1 ) n -1 n --
--
--
--
3. Using the method of difference, we obtain an= 3 - � for n � 3. Given that n 2009 , w have an > 2 + 2010 e 2 20091 3- > 2 + 32010 -:;;, 2010' yielding n > 4020. Hence the least value of m is 4021. for
n
�
14. Answer:
14
We have
8 + 8 sin 10° - 8 sin 10° -.8 sin 50° 9 + 8 sin 10° - 8 (2 sin 30° cos 20°) 9 + 8 sin 10° - 8(1 - 2 sin2 10° )
9 - 8 sin 50°
=
.·. v9 - 8 sin 506 Thus,
a = 1, b=
4 and
c=
16 sin2 106 + 8 sin 10° (1 + 4 sin 10° )2 1 + 4 sin 10°.
10, and hence
b+c
.
54
a
-
=
14.
+1
15. Answer: 3 It can be easily seen that the given equation has exactly one real r oot a, since (1) all polynomial equations of degree 5 has at least one r eal r oot, and (2) the function f(x) = x5- x3 + x- 2 is strictly increasing since f' (x) = 5x4- 3x2 + 1 > 0 for all r eal values of x. 1 It can also be checked that f(�) <0 and f(2) > 0, so that 2
a6 <4 since
a6 <4
-¢::::::} -¢::::::} -¢::::::} -¢::::::}
In addition, we claim that a6
a6
:::=::
:::=::
3
a4- a2 + 2a<4 a5 - a3 + 2a2 <4a 2a2 - 5a + 2 <0 1 2
3 since -¢::::::} -¢::::::} -¢::::::}
a4- a2 + 2a :::=:: 3 a5 - a3 + 2a2 - 3a 2a2 - 4a + 2 :::=:: 0,
:::=:: 0
the last inequality is always tr ue. Hence 3 :::; a6 <4, thereby showing that L a6J
=
3.
16. Answer: 8030
Any odd number greater than 1 can be written as 2k+1, where 2k+1 = (k+1)2-k2 . Hence all odd integers greater than 1 are not "cute" integers. Also, since 4m = (m+1)2-(m-1?, so that all integers of the for m 4m, where m rel="nofollow"> 1, ar e not "cute". We claim that all integer s of the form 4m + 2 ar e "cute". Suppose 4m + 2 ( for m:::=:: 1) is not "cute", then
for some integers positive integers x and y. However, x + y and x - y have the same parity, so that x - y and x + y must all be even since 4m + 2 is even. Hence 4m + 2 is divisible by 4, which is absurd. Thus, 4m + 2 is "cute". The first few "cute" integers are 1, 2, 4, 6, 10 For n > 3, the nth "cute" integer is 4n- 10. Thus, the 2010th "cute" integer is 4(2010)- 10 = 8030. ·
·
· .
17. Answer: 841
We have f(1) = 3, f(2) = 1, f(3) = 7, f( 4 )
= 36
and
f(x) = (x2 - x- 1)g(x) + (x- 1),
where g(x) is a polynomial in x of degree 3. Hence g(1) g(4) = 3. Thus
f(x)
Thus,
=
f(-1)
( X2 _ X _
[
-29,
so
-3, g(2)
=
0, g(3)
=
1, and
1). (-3) . ( x- 2)(x- 3)(x- 4) + (1). (x- 1)(x- 2)(x- 4)
]
(-1)(-2)(-3) (x- 1)(x- 2)(x- 3) · . + (3). . + (X _ 1) (3)(2)(1)
=
=
that its square is 841.
55
(2)(1)(-1)
•
18. Answer: 18
Solving for b, we get b = 10��t0�00. Since a and b are positive, we must have a > 100. Let a = 100 + m, where m is a positive integer. Thus b = lOO(lOO m)+lOO = 100 + 10!00. Therefore m must be a factor of 10100 = 101 x 2 2 x 5 2• Conversely, each factor r of 10�00) of the equation 100(a + m determines a unique solution (a, b) = (100 + r, 100 + b) = ab -100. There are 18 = (1 + 1)(2 + 1)(2 + 1) factors of 10100. Consequently there are 18 solutions of the given equation. In fact, these 18 solutions can be found to be (a, b) = (101, 10200), (102, 5150), (104, 2625), (105, 2120), (110, 1110), (120, 605), (125, 504), (150, 302), (200, 201), (201, 200), . (302, 150), (504, 125), (605, 120), (1110, 110), (2120, 105), (2625, 104), (5150, 102), (10200, 101).
�
19. Answer: 17 If both a, b are odd prime numbers, then p is even and p that p is prime. Thus a =2 or b=2.
�
4, contradicting the condition
Assume that a =2. Then b i= 2; otherwise, p =8 which is not prime. Thus b is an odd prime number. Let b
We shall show that k=1.
=
2k + 1, where k is an integer greater than 1. Thus
(
Suppose that k � 2. If k = 1 mod 3), then b > 3 and b = 2k + 1
=
(
O mod 3),
contradicting the condition that b is prime. If k ¢. 1(mo d 3), then
a contradiction too. Thus
k =1
and
b= 3.
Therefore
20. Answer: 10045 The number of grid points (x, y) inside the rectangle with four corner vertices (0, 0), (11, 0), (0, 2010) and (11, 2010) is (11 -1)
X
(2010 -1) =20090.
There are no grid points on the diagonal between (0, 0) and (11, 2010) excluding these two points, since for any point (x, y) on this segment between (0, 0) and (11, 2010), we have 2010x
y=--·-uBut for an integer x with 1 ::; x ::; 10,
0l�x is not an integer.
2
Hence the number of grid points (x, y) inside the triangle formed by corner points (0, 0), (0, 2010) and (11, 2010) is (11 -1)
X
(2010 -1)/2 56
=
20090/2 =10045.
(11, 2010),
For any grid point (x, y) inside the tr iangle formed by corner points we have
1
1
::::::
y ::::::
2009, 1
:S: X <
11y 2010'
(0, 0), (0, 2010)
and
Thus, for any fixed positive integer y, there ar e the number of grid points satisfying the is condition that :S: x <
.,i-�{0
11 l2 o :o J ' 1�{0 1 (0,2009. 0), (0, 2010) (11, 2010) 11 11 2 11 3 11 4 2009 . 11 l2010J l 2010 J l 2010 J l 2010 J l 2010 J 10045. 21. 2007 3. 3 3. 3. 3, 0. 3 3. as 2 is not an integer when :S: y :S: the tr iangle for med by cor ner points X
+
Thus the number of grid points (x, y) inside and is
X
+
X
+
+. . . +
X
Therefore the answer is
Answer:
Consider n distinct numbers where n 2: of fr iendly pairs is always n- for n 2: also n-
We shall show by induction that the number Hence the minimal number of fr iendly pairs is
If n = then there are no fr iendly pairs as each number is adjacent to the other two. Thus the number of fr iendly pairs is Assume that the number of fr iendly pairs for any arr angement of n distinct numbers on the cir cle is n- for some n 2: Consider n+ distinct number s on the cir cle. Letm be the largest number. Now consider the n numbers on cir cle after deletingm. The two numbers adjacent tom which originally for m a fr iendly pair do not for m a fr iendly pair anymore. Any other fr iendly pair r emains a fr iendly pair when m is deleted. On the other hand, any fr iendly pair aft er m is deleted was originally a fr iendly pair. By induction hypothesis, there are n- fr iendly pairs after m is deleted. Therefore, there are (n+ fr iendly pairs originally.
22.
Answer:
3
1) - 3
12345
1
We are given the equation f(x2+yf(z))
0 (1), 0 (1),
=
xf(x) +zf(y).
(1) 0. (2)
Substituting x = y = into we get zf(O) = f(O) for all real number z. Hence f(O) = Substituting y = into we get Similarly, substituting x = Substituting y =
1 (3) into
0 (1) in
xf(x) we get f(yf(z))
(3)
zf(1)
(4)
we get f(f(z))
for all r eal z. Now, using
zf(y).
(2) (4), and
=
we obtain
57
(5)
Substituting y =z =x into (3) also yields f(xf(x)) = xf(x).
(6)
Comparing (5) and (6), it follows that x2f(1) = xf(x), so that if x is non-zero, then f(x) =ex for some constant c. Since f(O) = 0, we also have f(x) =ex for all real values of x. Substituting this into (1), we have that c(x2 + cyz) = cx2 + cyz. This implies that c2 = c, forcing c = 0 or c = 1. Since f is non-zero, we must have c = 1, so that f(x) = x for all real values of x. Hence f(12345) =12345. 23. Answer: 86422 The number of ways of choosing r objects from n different types of objects of which + repetition is allowed is (n � 1 . In particular, if we write r-digit numbers using n digits allowing for repetitions with the condition that the digits appear in a non-increasing + order, there are (n �- 1 ways of doing so. Grouping the given numbers into different categories, we have the following tabulation. In order to track the enumeration of these elements, the cumulative sum is also computed:
)
)
+ n r (n �
Numbers Beginning with
Digits used other than the fixed part
1
1,0
2
5
2
2,1,0
3
5
3
3,2,1,0
4
5
4
4,3,2,1,0
5
5
5
5,4,3,2,1,0
6
5
6
6,5,4,3,2,1,0
7
5
7
7,6,5,4,3,2,1,0
8
5
From 800000 to
5,4,3,2,1,0
6
5
3,2,1,0
4
4
855555 From 860000 to 863333
-
1
)
�� =6 � �) =21 )
l!
0
I
5
27 83
=126
209
=252
11 =462 5 1 =792 o
6
=56
: �) 1 ;) c) (:) l
Cumulative
= 252
=35
461 923 1715 1967
2002
The next 6-digit numbers are: 864000,864100,864110,864111,864200,864210,864211,864220. Hence, the 2010th 6-digit number is 864220. Therefore, x =864220 so that l J =86422.
�
24. Answer: 309 LetS be the set of permutations of the six integers from 1 to 6. Then lSI=6! =720. Define P( i) to be the subset of S such that the digit i + 1 follows immediately i, i =1,2,3,4,5. 58
Then =
� IP(i)l (�)5! _I; 1P(i1)nP(i2)1 2
21 <22
(� )4!
=
. �. 1P(i1) nP(i2) nP(i3)1 (� )3! . _I; 1P(i1)nP(i2)nP(i3)nP(i4)1 (!) 2! . IP(1)nP(2)nP(3)nP(4)nP(5)1 (�) 1!.
21 <22<23
=
=
21 <22<23<24
=
By the Principle of Inclusion and Exclusion, the r equired number is
25. Answer: 2011 Note that A =
(COo10) - (2��0))2+ ((2010)- (2010))2+ ((2010) - C010)) 2 2 0 1 1 +...+ (G���) G���) ) 2 � f ((2010k ) _ (���))2 . k=O =
_
Obser ve that
is the coefficient of
x2010 in the expansion of the following expression:
We also have
2011 ((2010) - ( 2010 )) k 2011 (2010) k - 2011 ( 2010 ) k L L k L k k 1 k 1 k=O k=O k=O (x+1)2010 _ x(x+1)2010 (1 _ x)(x+1)2010 2010 ((2010)- ( 2010 )) k 2010 (2010)xk-! 2009 (2010 )xk+1 + L1 k k 1 x k=O L k k 1 L1 + k=k=(x+1)2010 _-(1 x+1)2010 (1 _ 1/x)(x+1)2010. 59 X =
=
and
X
=
=
=
X
X
=
X
The coefficient of x2010 in the expansion of the following expression: (1 - x)(x + 1)2010 (1 - 1/x)(x + 1)2010 = (2 - 1/x - x)(x + 1)4020 is 2 Hence
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4020 2010
- 4020 2011
A=
- 4020 2009 4020 2010
Consider the inequality:
sA 2:
=
_
2
4020 2009 ·
4020 2010 '
s(1 - 2010/2011) s
2:
Hence the answer is 2011.
60
4020 - 4020 2 2010 2009 .
2011.
2:
1,
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO)
2010
(Open Section, Round 2) Saturday, 26 June 2010
0900-1330
INSTRUCTIONS TO CONTESTANTS
1.
2. 3.
4.
Answer ALL 5 questions. Show all the steps in your working. Each question carries
No
10
mark.
calculators are allowed.
1.
Let CD be a chord of a circle r1 and AB a diameter of r1 perpendicular to CD at N with AN> NB. A circle r2 centred at C with radius CN intersects r1 at points P and Q. The line PQ intersects CD at M and AC at K; and the extension of NK meets r2 at L. Prove that PQ is perpendicular to AL.
2.
Let an, bn, n = 1, 2, . . . be two sequences of integers defined by a1 for n;::: 1,
an+1 bn+1
=
=
=
1, b1
=
0
and
7 an + 12bn + 6 4an + 7bn + 3.
Prove that a� is the difference of two consecutive cubes. 3.
4.
5.
Suppose that a1, ..., a15 are prime numbers forming an arithmetic progression with common differenced> 0. If a1> 15, prove thatd> 30, 000. Let n be a positive integer. Find the smallest positive integer k with the property that for any colouring of the squares of a 2n x k chessboard with n colours, there are 2 columns and 2 rows such that the 4 squares in their intersections have the same colour. Let p be a prime number and let x, y, z be positive integers so that 0 < x < y < z < p. Suppose that x3, y3 and z3 have the same remainder when divided by p, 2 show that x + y2 + z2 is divisible by x + y + z.
61
Singapore Mathematical Society Singapore Mathematical Olympiad (SMO) 2010 (Open Section, Round
2
solutions)
Extend DC meeting r2 at H. Let the radius of r2 be r. Note that DN = NC = CH =r. Since M is of equal power with respect to r1 and r2. Thus MN MH = MC ·MD. That is MN(MC + r) =MC(MN + r) giving MN=MG. Thus M is the midpoint of NC. 1.
·
A
As K lies on the radical axis of r1 and r2, the points C, N, A, L are concyclic. Thus LALC=LANG=goo so that AL is tangent to r2. It.follows that AC is perpendicular to NL at K, and hence MN=MC=MK. Now let PQ intersect AL at T. We have LTAK = LKNM = LNKM = LLKT and similarly LTLK = LAKT. Consequently, 2LKTL = 2(LTAK + LAKT) LTAK + LAKT + LLKT + LTLK= 180°, which means LKTL=goo. First we shall prove that a� is the difference of two consecutive cubes. To do so, we shall prove by induction that a� = (bn + 1)3- b�. When n = 1, this is true. Suppose for n 2: 1, this is true. We have 2.
(bn+l + 1)3- b�+l =3b;+l + 3bn+l + 1 =3(4an + 7bn + 3)2 + 3(4an + 7bn + 3) + 1 =48a; + 147b; + 168anbn + 84an + 147bn + 37 =(7an + 12bn + 6)2 + (3b; + 3bn + 1)- a; a;+1 + (bn + 1)3- b�- a�=a;+l =
where the last equality is by induction hypothesis.
62
Lemma: Suppose p is prime and a1, ... , ap are primes forming an A. P. with common difference d. If a1 > p, we claim that p I d.
3.
Proof Since p is prime and every
ai is a prime > p, p does not divide ai for any
the pigeonhole principle, there exist 1 � i < j � p so that ai aj aj - ai = (j -i)d, and p does not divide j - i. So p must divide d.
( mod p) .
i. By Now
Apply the Lemma to the sequences at, ... , ak fork= 2,3, 5, 7,11 and 13. Then all such k's are factors of d. So d > 2 3 5 7 11 13 > 30, 000. ·
4.
·
·
·
·
The answer is 2n2 -n + 1.
Consider an n-colouring of the 2n x k chessboard. A vertical-pair is a pair of squares in the same column that are coloured the same. In every column there are at least n vertical-pairs. Let P be the total number of vertical-pairs and Pi be the number of vertical-pairs with colour i. Then P = P1+ + Pn � nk. Thus there is colour i with Pi � k. There are = 2n2 -n pairs of rows. Thus if k � 2n2 -n + 1, there is a pair of rows that contains two vertical-pairs with colour i.
e;)
·
·
·
Next for k = 2n2 - n, exhibit an n-colouring where no such sets of 4 squares exists. Note that it suffices to find such ann-colouring for the 2n x (2n-1) board. We can then rotate the colours to obtain n of these boards which can then be put together to obtain the requiring n-colouring of the 2n x (2n2 -n) board. For each i = 1,2,...,2n- 1, let Ai = {(i,2n-1+i),(i+1,2n-2+i),...,(n-1+i,n+i)}, where 2n+k, k > 0, is taken to bek. Note that the pairs in each Ai give a partition of {1,2,...,2n}. Moreover, each pair of elements appears in exactly one Ai. Now colour the squares of column i using n colours so that the two squares in each pair of Ai receive the same colour and the colours the 2n pairs are mutually distinct. This gives ann-colouring of the 2n x 2n -1 board with the required property and we are done.
5.
It is given that
p I x3-y3= (x -y)(x2 + xy + y2).
Moreover,-p < x-y < Oand sop f x-y. Thusp I x2+xy+y2. Similarly,p I y2+yz+z2 and p I x2 + xz + z2. Hence
p I x2 + xy -yz -z2 = (x -z)(x + y + z). Since p f x -z,p I x + y + z. Note that 0 < x + y + z < 3p. So x + y + z = p or 2p. We will show that p I x2 + y2 + z2. Assuming this for the moment, the proof is complete if x+y+z = p. If x+y+z = 2p, then x+y+z is even and hence x2+y2+z2 is even. So both 2 and p divide x2 + y2 + z2. Moreover, p > 2 and so 2 and p are relatively prime. Thus 2p divides x2 + y2 + z2 and the proof is also complete in this case.
63
Fg{^QZ?QcZvQgUQr FUgZ Qmwf`zy{dN`~gds`~ma`qPrszm`cQNP F<
-( ʥŭʥ
@ ʥ ʥȩʥ -ʥ
#"$&!&)
,371:BMMo '/o 0534#.,3 ,41:9.51:,3713:.,:4":,371:3"4:/1.6#: .1:4":*5)4&):".#:0534#.,3 : ,41:9.51:,3713:.,:4":,371:3"4:9:3"#,!:4": 5):.,4#,#,!:4":)441: : Fo Go =ʥ.1::.23/.,#,!:4.:4":.114:,371 .1: 4": .4"1: 3".14: 0534#.-3 : 71#4: 9.51: ,371: #,: 4": ,371: 3"4: ,: 3": 4": //1./1#4:5): ).7:9.51:,371 : .:34/3:1:,:4.:'534$ :9.51:,3713: ":0534#.,: 11#3: o *1( .:)5)4.13:1:)).7:
)))) ) )) ) ))))
ʥ
'&#)")'%&"!%)
;
iʥ4dʥ ʥ4_ʥʥ))ʥ)Bʥ*ʥ )ʥ#bs ] ʥ ʥʥ]ʥ ʥ$B ʥ ʥ ʥ*ʥ ]ʥ*]#ʥA)ʥ]ʥʥ*ʥ)Bʥ*ʥ )ʥ "ʥbs )ʥ])ʥ$B ʥ] ʥ ʥ ʥ*ʥ*]ʥ14; ʥ1 ";
%aʥ %Fʥ %Yʥ %jʥ ;
ʥ ʥ "ʥ +ʥ *ʥ*ʥ]ʥ *&ʥ
Uʥ]ʥ) ʥB*ʥ]ʥ )ʥ*ʥ< ʥt =ʥ)ʥʥ ʥ]ʥ )ʥ*ʥ< ʥ t94ʥ)ʥ+ʥ)&ʥ] ʥ©4t =; ʥ ß7ʥɐʥ]ʥ ʥ*ʥ]ʥ] ʥ)* 94=#ʥ 1;
ʥ ";
%aʥ %Fʥ %Yʥ %jʥ &o
Êʥ
Êʥ
£Êʥ
"Êʥ *ʥ*ʥ]ʥ *&ʥ
iʥkʥ ʥ ʥŀʥA)ʥ]ʥ ^)ʥ& Bʥ*ʥk]ʥ] ʥ]ʥ**) )< B);ʥ]*BDʥ
k& T&
#
%aʥ ʥ %Fʥ %Yʥ %ȡʥ Ǒʥ
ʥ
*o
@ŭʥʥʥ ʥ ʥ ʥ ʥʥŰʥ$&ʥʥ ƪʥ 3ʥ ʥ4ʥʥ ʥ®ʥUʥʥʥVXoʥ ʥʥʥʂʥ2ʥÖʥʥʥʥW go
T&
%aʥ %Fʥ %Yʥ %jʥ .o
ʥ ʥ ʥ "ʥ +ʥ
:;; s!ʥ 2 o Ä&ʥʥ ʥʥʥ< Źʥ ʥ ʥ ʥ!ʥ ʥʥeo ʥʥʥʥ Ä&ʥʥ Äʥ
! Öʥʥ ^ʥ& ʥʥ: HO
T&
%aʥ %Fʥ %Yʥ %jʥ 0o
( ʥ P-ʥ
(£ʥ mʥ ʥ &ʥ
@ʥ<ʥ/6; Ãʥ/6;*6;:5; ʥ/8; ; -ʥAʥ/98 T&
%aʥ %Fʥ %Yʥ %jʥ
ʥ - £ʥ -ʥ £ -ʥ ££-ʥ
ʥ
3o
Fʥʥʥ ʥŻʥ ʥ ʥʥ ʥ&ʥ ʥ ʥʥ´ðʥ$ ʥʥ ʥ ʥʥʥ^ʥʥʥAʥʥ$ ʥʥʥ$ ʥ ʥʥ &ʥʥʥʥ
T&
)' To^o
%aʥ U Fʥ #To %Yʥ #Uo %jʥ #T 7o
&ʥ ʥbo 6ʥ ʥ ʥʥ Ãʥʥųųɭʥ< Dʥ
beo6 ʥ ;)+
bo
ʥ b!o ' n
eʥʥ& ʥʥ Gbo6 ʥ #G T&
%aʥ Fʥ %Yʥ %jʥ
-
#G
)" )G ʥ ʥH
&ʥ ʥ6ʥ
n
T&
%aʥ Fʥ %Yʥ %jʥ
b #!( #boʥbo#´ðʥʥʥ& ʥ 6
P-ʥ
#
;
@ʥʥʥ$LLgʥL ʥLʥ0; h+ 3; : LÃL ,ʥ : ,ʥ; J LʥʥLLʥ ʥ/;;h;3; ;:Lʥ T&
# #
Yʥ
#
"ʥ
# #
ʥ
"$&)'%&"!%) ;
@ʥ ʥÝʥʥʥiFYʥgLLʥʥ ʥŝʥLʥ ʥį ʥLįʥ ʥʥ LʥʥʥLʥ ʥƛ ʥcƚʥLʥʥ@ʥ Lʥʥʥʥʥʥ Lʥʥʥ ʥʥLʥ#0 7ʥLʥ ʥʥʥLʥʥʥLʥ "ʥ/ @gLLʥʥ ʥ ʥʥʅʥʥʥɠʥL 2ʥʥʥʥ gʥʥ ʥL ʥɗʥʥ ʥʥUʥʥ Lʥʥʥ ʥʥʥ ʥ ʥʥ ʥʥʥLʥʥLʥ!; / ʥɏʥʥg ʥʥ!
;
@ʥL ʥʥʥ ʥ$ ʥ ʥ$ʥʥLʥ!; ʥLʥ9|ʥ ʥ ʥ ʥ ʥ$ʥ ɳʥLʥ9ʥ ʥLʥ Z ʥg $ʥ @ʥg $$ʥ Lʥ9ʥLʥ "ʥʥʥ$ ʥ ʥʥʥ ʥUʥʥ LʥʥLʥ!; LʥLʥʥ2ʥ´ðʥʥg ʥʥ ʥ$ ʥʥLʥ4ʥʥʥʃ ʥLʥ+: ,: +,:
!;
9ʥ 1;
£ʥ
%o
*o
.o
-ʥġʥ ʥ Ùʥ ʥʥʥ*$ʥʥġʥ*ʥ ʥ<*ʥAʥ ʥʥʥ*ʥ ʥ$ʥ<ʥ*ʥ ʥ*ʥ$ʥʥ ʥ ʥʥ ʥ ʥ ʥʥ*ʥ ;ʥ*ʥ ʥʥ*;ʥ ʥ"ʥʥʥ ʥ*ʥʥ ʥʥ ʥʥ ñ ʥ*ʥ*ʥʥʥ*;ʥ *ʥʥ ñ ʥ*ʥ*ʥ*;ʥʥ#y [o Aʥʥ & ʥ*ʥ[
3
!#$; ʥ ʥ ʥ ʥʥ,#!$; -ʕ®ʥAo < ʥʥ*ʥ*ʥʄʥ ʥ!#; ʥ#$; ʥ*#ʥʥ ʥ*ʥʥ< ʥ!#%'; ʥŋʥ>ʥ ʥʥ ʥ*ʥ ʥ< ʥ#$)+;ʥ "ʥ>#ʥAʥʥɁ ʥ*ʥ ʥ%#+;ʥ>ʥ
+;
:
';
0o
1;
!;
ʥ
ʥ '
ʥ rʥ ʥVu *ʥ ʥY Y ʥ'ʥY ʥ ' ' ' ʥY ʥ ʥY ʥY # 888 ʥY ʥY ʥ ʥY ʥY "ʥRV 5ʥ ʥʥ*ʥǝʥAʥʥ& ʥ*ʥj'5ʥ
j
4o
7o
8o
&ʥ ʥg'
ʥ; h'h ( ʥ ʥg1 h,ʥ ʥ 2ʥ´ʥʥ& ʥ*ʥgh0 g'h8
Uʥ p !ʥF''6ʥ3 ʥ ʥ l* , !ʥ;'ƫʥÖʥʥ& ʥ* !ʥ,ʥ' -r ʥÚʥ ʥÈʥ$ʥ*ʥ*&ʥ$ʥʥA*ʥ*ʥɌɟ^ʥʥ, ʥʥ ,:( 2ʥ5ʥ( -2ʥŋʥ( 52ʥ'5ʥ( Ĥʥ $ʥ ʑ2ʥ Ú2ʥÈ2ʥ ÈĤ2ʥʥ*ʥ ;ʥʥ ʥ*#ʥAʥʥ& ʥ*ʥ5®ʥ
"ʥ
o
Acʥ ʥ ʥ ʥʥ $ ʥ ʥ$&ʥ c
ʥ ơ!2ʥ6ʥ ʥ ɓʥ ʥ < ʥ !ʥ$' 6ʥ' -ʥP -!
o
-6ʥP -ʥ0ʥ#
Aʥʥʥ$ ʥʥ
ʥ
ʥ ' ʥ
ʥ
ʥ
ʥ
ʥ
ʥ ʥ,ʥ ʥ' "ʥ,ʥ 'ʥ' +ʥ,ʥ -ʥ
o
@ʥ ʥBʥʥ ʥ Bʥ 3Òđ2ʥʥʥ ʥʥ ¤=2ʥ3¤TX2ʥ=TÀđ ʥX ¦§ʥ ʥ ʥ_2ʥʥ_2ʥ+ ʥ_ʥ ʥ ʥ_ʥ$&B;ʥAʥʥ ʥ §Àś _ʞʥ 3ʥ
ʥ =ʥ
¦ʥ
đʥ
%o
*o
śʥ
Òʥ
j& B ʥ ''ʥ( ʥżʥ ,ʥ ''ʥ,ʥ ʥżʥ ʥ Aʥʥʥʥʥʥʥʥʥ ʥ 2ʥ2ʥ###ʥ2ʥ -Ĥʥʥʥʥʥʥ ʥ
ʥ 'ʥ
.o
S&ʥ ʥ ʥ ʥ.ɵʥSʥ ʥSʥ
3ʥJʥ ʥ###ʥʥ5# #ʥ Aʥʥ& BGʥ 5ʥʆʥ ʥ.ŷʥ,ʥ._ʥ,ʥ. ʼnʥ,ʥ.~ʥ,ʥ
.ʥ zƽdʥ0ʥ"P '.z ă_ʥ
!!!O
1o
ìʥʥS ʥq 3ʥSʥ ʥS BʥSʥȮ q3ʥ3?ßʥ ʥq3ʥ Jʥï#ʥvʥo ʥ ʥ ʥ $Sʥʥ 3ʥGʥ ʥ " vʥ vʥ3JJʥ3#ʥìʥʥ&ğS BʥSʥʥvʥ3ʥï ×ʥʥ ʥʥʥS Bʥqvʥʥ_#ʥ 0O
o
ʥ
EEEEEEEEEEEEEEEFGGOCCCCODO
ʥ
qʥ 3o
ļ
ʥ
3ʥ
Jʥïʥ
iʥ[dʥ!_ʥ!\ʥ[~ʥʥʥGʥʥʥʥ
7o
@ʥSʥq 3ʥq ¤ʥ ʥq3¤ʥʥ ʥ ʥq 3¤ʥ ʥS BʥS Bʥ S##ʥ©¾q3ʥ3©¾q¤ʥ3©3q¤ʥ3?ß#ʥS&ʥ ʥt ʥ3'ʥq3ʥ3 ʥ ʥq¤ "ʥ´ðʥ ʥ& BGʥʥ H ʥʥėĘq 3_ʥH ʥʥėĘq Î _ʥ, H ʥʥėĘq3ďĀ_ʥH ʥʥ3Î _#ʥ 0O
8o
5ʥĂJ ASʥ ʥ B ʥ$SS&ʥSʥ5ʥ ʥSʥ ?5ʥ ,ʥJJʥ S#ʥ +ʥ
QU
ʥļŻʥGSBʥ
%o
Aʥ8ʥ& BG8ʥoʥ8ʥ BB8ʥ$&8ʥ88ʥOʥGʥ ʥ8ʥ8
!
,ʥ Oʥʥ ʥx , JssOʥ-ʥ ʥ3
ʥB;ʥ88ʥBG#ʥ %o
% o
%%o
&+o
%.o
Uʥ ʥ B8ʥ!#$; 8ʥB8ʥoʥ8ʥ BG8ʥ!%; ʥ#(; 8ʥʥ ʥJ ʥ8$8&8B;#ʥ Aʥ8ʥB 8ʥ$B8ʥ88ʥ& BG8ʥoʥ8ʥB8ʥoʥ8ʥʥ Bġ8ʥ$); T oGʥʥG8ʥʥoʥÝʥʥ$ ʥoʥ8$8 8ʥʥoʥ8^ $B8ʥ JJ2ʥ " "ʥ ʥ''#ʥAʥ8ʥ BʥG8ʥoʥGʥ$B8ʥG8ʥ ʥ 8ʥ &B8ʥ;ʥ'ʥʥJJʥGʥʥ#ʥ
Oʥ ʥ5ʥ 8ʥʥ$&8ʥc88ʥ ɍʥJʥċČčʥOʥċČčʥ5ʥċČčʥ#ʥAʥ8ʥG8ʥoʥ $ ʥoʥO2ʥ5ʥGʥ ʥ8ʥ$GʥO5ʥʥ&B8ʥ;ʥ#ʥ ĕʥ8ʥʥ ʥJ2ʥ 2ʥʥ ʥ2ʥ×ʥ8ʥG8ʥoʥJʥ8
-ʥ
Fg{^QZ@QcZvQgUQr FUgZ Rmwf`zy{dO`~hdt`~ma`rPrtzm`cROP F<
HōʥF|ʥ Hʥʥʥʥ ʥBʥʥ ʥ"ʥBʥ ʥ ʥ ʥʥʥ$ ʥ1
FG V g q r s u
BC
(
)*+ + HIJ DE
} ~ t v
w x y z U V
v t t }
KLMNOP =>?@ABO "O
o
HȊʥHʥ Ş ʥɡǡ 3ʥ-ßʥ BÙʥ ʥʥ$ʥ ; ʥ&B $ʥʥ %$ 1;
1;
;
ʥ
ʥ"ɶɅ( "Êʥ0ʥ Ê# ŏ#ʥV ʥ 0ʥ8
ʥ
3ʥ
%o
÷µÛ¸ʥ$ _ʥ
ʥ 0ʥʥʥ ʥ !( '(!ʥ 0ʥ ʥʥ ʥ "# ( !(#Ø Zʥ ^)ʥ& ʥ*ʥyʥ1ʥ ʥʥ ʥ ʥ 0 ʥ *o
÷ܸʥB% iʥvʥʥʥ)$*)ʥ*ʥĖʥ ʥʥʥʥ$*)ʥÜʥʥ)ʥVYoʥʥ)ʥ )#ʥ fo
@ʥ;ʥ)) ʥſ ʥ0ʥJ ʔ ʥv3ʥ0ʥJ ʥZĖ0ʥ"o0ʥ ʥ _ ( J _ʥ0ʥ .o
÷µÛuʥ# Jʥ Jʥ 61ʥss!ʥ ƾ! ! 0 ʥJ( 6ʥJ h * 6ʥJĎʥJ (s h6 hs# V)ʥ* !ʥ 6ʥ ʥ) 6ʥʥJʥ)ʥ ʥ *ʥ*ʥJ ʥ)ʥ6ʥ0ʥ(- ʥ(+2ʥ(" ʥ(#ʥ @ʥ ^)ʥ& ʥ* 6ʥ)ʥ(ʥ
0o
÷ܸʥ ; ; ʥ J iʥhy0ʥ]y0 ]y2? 0ʥ5 _ʥ Mhy 1ʥ M5 _ʥ1ʥJʥ JJʥ0ʥ - # ;
yS
ySu
ēʥʥ*ʥ ʥ Hhy 1ʥ/;8; /8; ʥ/;8;1ʥ - ʥʥ/8;0ʥ-Ł yS
J ʥ
3o
Huʥ
FSʥʥ<SB ʁ BʥS Bʥʥ;ʥñSʥʥʥʥʥ ñ ʥ#ʥ ìʥʥ ;ʥʥʥ ʥʥ<ʥ$ ʥSʥ&ʥ;ʥʥ$ ʥʥʥ Bʥ&ʥ;ʥʥʥʥ ; > (
ʥ ɬĨ# ʥ &ʥ!
>ʥy Ĩʥ1ʥ !
7o
Huʥjʥ K>ʥ,ʥǤʥ1ʥ"ʥ#Kʥ,ʥ6>ʥ0ʥ ʥ,ʥ"ʥ Vʥ Kʥ,ʥʥKʥ,ʥ6>ƺ "ģʥ,ʥ 6ʥ0ʥ ʥ,ʥĢ > > :; K , 6 , K , 6 , 1
, "ĢʥéK , 6 , 1 , Ģɋ
8o
HDʥjʥ 6ʥ1ʥK> P ">ʥK> P >ʥ1ʥK~ P K>ʥ,ʥ ǿ"ʥ1ʥƞʌ> ( "ƣ> ¬ - VʥĝęSGʥ& BGʥʥ6ʥ1P -#ʥ
o
HuʥƢHʥ g, h+ 3; x1ʥ 77'7 7ʥ 77+7 7ʥ 77-7 +7ʥ 77 7 7ʥ 777 7ʥ 77"7 #ʥ
o
Huʥ'ʥ
"Kʥ ʥ V ʥ
ʥ
ʥ'ʥ ʥ H ʥʥV ʥ >ʥ "Kʥ>ʥ,ʥ-Kʥ>ʥ1ʥ ʥ>ʥȚʥ'Kʥ1ʥ {ʥ .#ʥH ʥ S¡ "K-K¡ ʥ{
ʥ
o
M
Dʥʥ'ʥ ʥ iʥʥ ʥʥʥÇ ʥ ʥʥ ʥʥvʥ ʥĔʥÇg;#ʥ ʥ v1 ʥĔ#ʥV i1 ʥvȀ'ʥĔʥ1 "ʥ ʥi 5ʥ1ʥ'# ʥ gʥ "ʥ ,ʥ Ŀʥ NJNj ǭʥ Ŀ DZʥ
%o
M
Dʥʥ ʥ ȵʥʥʥ ʥÇʥʥ Ȇʥ,ʥ ʥ1ʥ #ʥ
*o
MDʥ ʥ Aʥʥ ʥʥ"ʥʥʥ"Ħʥ ;#ʥm^2ʥʥ ʥ'ʥÇ ʥʥʥ"ʥ ʥʥʥʥʥ;#ʥZʥk 3 ȂFĈʥy Ħʥ1ʥ #
.o
MDʥ"ʥ ʥ ʥȝ3 4ʥ-ßʥ ɣʥ ʥʥÇʥ3ʥʥʥȜ34Įʥ¦#ʥ "(+ʥ1 ¦ʥ Xʥ
³ʥ
0o
MDʥ ȁʥ
ʥ
1;
;
ʥ
ʥ
ʥ Ą 1ʥ mʥ ʥ 5ʥy 5ʥ,ʥ ʥ ʥ 5ʥw 5ʥʥ 5ʥ,ʥ ʥy 5ʥ ʥ ʥ #ʥ y 5ʥ,ʥ ʥ
ʥ
ʥ
ʥ # Zʥ ʥY Y ʥ,ʥʥY ʥ , ,ʥ ,ʥ ,ʥ
Y ǬʥY ʥ ʥY ʥY " ʥY ʥ ʥY ʥY "ʥ
ʥ
ʥ
ʥ ²ʥǰʥ ʥY ʥ( ʥY "ʥ 1ʥ "ʥ#
;
;
;
ʥ
;
3o
M¸ʥ ʥ
ʥ( ʥ¹º ( ʥ
iʥKʥ0ʥ.ʥʥ 7ʥ6ʥ0ʥbʥā ʥKʥ( 6ʥ 7ʥʥKʥ( ʥʥĄX ʥ;6ʥʥ ʥʥƻ6ʥ »ʥKʥʥX Wʥ6ʥĄ6ʥ
; :;Wʥʥ¹º»ʥʥ0ʥ
¹º»ʥK(6ʥ ( 7o
123
.b (.ʥbʥ0ʥ #ʥ
Muʥ ʥ V$$ʥ6ʥno #ʥAʥ¨ʥ6ʥ* 6ʥʥK ʥ ; 'ʥʥ gʥKʥ0ʥ'ʥ ʥʥ GK¨ʥKʥʥ6ʥ0ʥ ʥ
ʥ g 6 0Pʥ ; hʥ]o "7 Vʍhʥ] V$$ʥKʥaoʥAʥ GK¨ʥKʥʥ6ʥ0ʥ ʥʥ gʥ6ʥ0ʥĎʥʥ !7 V !Ďʥ#
ʥ ʥ Ńʥ Zʥʥʥʥʥ< Źʥʥ ( 6ʥʥK ʥ ;'ʥ ʥʥ Kʥʥ6ʥ0ʥ ʥ¹º»ʥKʥ0ʥI 6ʥW Ă7
8o
Huʥ'ʥ V)ʥÚʥʥÈʥ0ʥ"5( '7 ÚʥʥÈʥ0ʥ, -ʥ)ʥ)ʥʥV Ú 0ʥ 7ʥÈʥ0ʥʥ ʥ,:Wʥ'®ʥ
o
Huʥʥ"ʥ mʥ ʥ Kʥʥ6ʥIʥ -ʥ( -KʥP -6ʥ¡ 0ʥIʥ -ʥ % -ʥ 7ʥʥK6ʥ0ʥ -®ʥ ȉʥK7 6ʥWʥ 7ʥ -7ʥ'7ʥ +'7ʥ 7ʥ-7ʥ-7ʥ 7ʥ +'7ʥ'7ʥ -7ʥ ®ʥ
o
Muʥ +"ʥ
ʥ
ʥ
ʥ
ʥ
ʥ
ʥ
ʥ iʥÒʥ0ʥ ʥ ʥ ʥʥ ʥʥ "ʥʥ 'ʥʥ +ʥʥ -ʥ#
ʥhʥL ʥhʥ +'Ń 'ʥ hʥÒʥhʥ 'ʥ 789 +"ʥ< F ʥ -ʥ ʥ
o
HDʥ "ʥ iÙʥ ʥ ʥʥʥ ʥʥ ʥ 3Òȧ7ʥʥ gʥ 3Ñʥ0ʥ 3=ʥʥ=¦Ȩʥ ¹ºƠ» ¾3ďĀʥ34¥Ŗʥʥ¥X §Śʥʥ§ÀÑʥ0 3ďĀʥ 4=ʥ=¥Àǚħʥ¥X §ŚʥʥX¦ŗʥ X ¦ŗʥ 456 34¥Ŗʥʥ§ÀÑʥ0ʥ 4=ʥ =¥řǛħʥ ¹º»ʥʥʥ§ÀÑʥ0ʥ ʥʥ+ ʥʥ ʥ 789 §řÑʥWʥ "®ʥ
ʥ
%o
ÓÔµʉDʥ'ʥ i Â1ʥ ''ʥ¬ ʥî1ʥ ''ʥ 1 ʥ Âʥîʥ >ô ʥ;-ʥ \ʥ1 ÂʥȺ\ʥ1ʥÂ\ʥʥȹ\ʥʥȷîÂʥʥȻ V)ʥÂ\ʥʥî\ʥ1ʥ Ėʥ1ʥ \ʥ1ʥ ʥʥ ' ʥèʥ \ ( ' ( ʥ;ʥè ( ' >ʥʥ' ʥʥ ʥ1ʥʥ 1ʥ'# :;
*o
ÓÔܸʥ ʥ Áʥ)ʥʥʥʥ)uʥ ʥ ʥ ---ʥÁʥʥʥʥʥ )gʥ)ʥʥ.ʥʥ:ʥ`1ʥ
ʥ dd \ d d\ Lo )ʥ ã ă F\Ŧdʥ1ʥ F>ʥ1ʥ'+ʥmʥ ʥ
ʥʥʥʥ$)ʥ ʥ
ʥʥ Vʥʥʥʥ)ʥ)ʥ'+ʥ( ʥ( "ʥ1ʥ"-ʥ ʥ ʥ ---ʥÁʥʥʥʥʥɫ )gʥ)ʥʥ.ʥhʥa1ʥ ʥ Uʥ)ʥdǧã\ădF\Ŧdʥ1ʥd>F>ʥ1ʥ""ʥmʥ ʥ ʥʥʥʥ$)ʥ ʥ ʥVʥ ʥʥ)ʥ)ʥ""( ʥ1ʥ"ʥ ʥ ʥ -ʥÁʥʥ ʥʥ -ʥ )´ʥʥ$$ʥ NO
NO
ŏʥ@ʥ ʥʥʥ)ʥ )Ã)ʀʥ$$ʥ)ʥ"-ʥʥ"ʥʥ ʥ1ʥ ʥ
.o
ÓÔµÛDʥʥʥ ĭ .z1ʥ ʥʥ >ʥʥ \ʥʥʥ zʥ1ʥ zãȤ ĸ i !ʥ1ʥ ʥµÛʥ gʥ.ŷʥʥ.dʥ.ʥ >ʥʥʥ ʥ . 1 H .d1 ʥʥ51ʥ ʥ z "#"#ʥ 5ʥ "#ʥ ʥ"ʥ( ʥ ʥ ʥ ʥ ʥPd zãʥdʥ ʥ#"# ʥ ;O
0o
ÓÔµʊDʥ "ʥ
ʥôʥʥnʥw ʥbh; "ʥ> mʥ ʥ ʥʥſ tvʥ1ʥ2
3o
ÓÔÜDʥ'ʥ Aʥ)gʥ< )ʥ!dʥʥ!>ʥʥ[\ʥʥ[~ʥ1ʥʥ1ʥ!dʥ!>ʥʋ\ʥʥ!~ʥʥ[\ʥ[~ʥ!ɞʥʥ!>ʥ1ʥP-ʥ !dʥ!>[\[~ʥ1ʥ( -ʥmµÛʥ!dʥ!>ʥ1ʥ-ʥèʥ[\[~ʥ1ʥP ʥVʥ!dʥʥ!>ʥ1ʥ ʥ!\ʥ[ʥ ~ʥ1ʥ( ʥ Zʥyʥ² [øʥ[>ʥ[ʥ øʥ[\ʥ[øʥ[~ʥ[ʥ >ʥ[\ʥ[ʥ >ʥ[~ʥ[ʥ \ʥ[~ʥ1ʥ-ʥʥ ʥY ʥ( ʥ( ʥ1ʥ'#ʥ
ʥ
7o
Muʥ ʥ m*ʥ ʥM *«t 9>pʥM ʥ*ʥ«t 4>ʥM ʥ*ʥ«t9ë >ʥM ʥ*9ë >ʝ V*ʥM ʥ*ʥ«t 9>ʥ, M ʥ*ʥ«t Î >ʥM ʥ*ʥ«t9ë >ʥM ʥ*ʥ9Î>ʥ ʥw M ʥ*ʥ«t ë >ʥM ʥ*ʥ«t94>ʥ, M ʥ*ʥ9Î>ģ Wʥ ʥY ʥ ć ʥY 'ʥY "eʥ ć ʥY 'ʥY >ʥʥ ć ʥY ʥY "Øģʥ # n
n
8o
M
Dʥ
ʥ Ǚʥ -l N N ʥWʥ-ʥʥ ʥ #ʥÐʥ 5P ʥ )ʥ ʥ*ƹź*ʥ)Nʥ )*7 F*ɝʥ lʥ ʥ l ʥ -lʥʥ
ʥ ʥlʥ ʥ)ʥ N*ʥ ʥ*ź*ʥ)Nʥ )*ʥèʥi ʥ$*))&ʥ)ʥlʥWʥ
# :O
%o
:O
MDʥ-ʥ K>ʥʥ Oʥ,ʥʥKʥ,ʥ NssO ʥ-ʥ ʥK Kʥ POʥ,ʥʥio OP ØP - #ʥ@) ;)Nʥ)ʥ*N)*ʥ)ʥ ʥ*N;ʥ)ʥO( Ø( -ʥ)ʥ ʥ$ʥ< 7ʥ ;ʥl >ʥ ZʥO( ØPl >ʥWʥ -ʥWʥ'>ʥw ʥéʥ n
n
¨ʥO (ňʥ,ʥlʥpʥ -ʥ ʥ¨ʥO (ň(lʥWʥ 7ʥ*ʥ ò O( ò51 +' ʥò OP òPlp '7 *ʥ ¨ʥOʥP ʥ¨ʥʥ,p : ʥ-ʥɀʥ¨ʥOʥ( ʥ) lʥWʥ # V*N&c2ʥlʥWʥʥjo 7ʥʥjo '7ʥʥioʥé@ʥ NNʥ$*))&ʥlʥ)ʥ-#ʥ
%o
Muʥʥ M ʥ*ʥʥ) Nʥ) w =ʥw 94 Wʥw 4ʥw 9³ W w 9ʥw 4X V)ʥ =pʥʥ ʥ9³ Wʥ 7ʥ94 Dʥ 4ʥ ʥDʥ ǜʥi 4ʥ K7ʥʥ94ʥ Kʥ n
n
n
ĕ)ʥ@) NʥÐ< N);7ʥ 9ʥhʥ 4ʥʥ94ʥ ʥ94ʥhʥ 9ʥʥ 4ʥșʥ Kʥhʥ 9ʥhʥK#ʥ A*ʥ4Xʥpʥ
ʥhʥ4Xʥhʥ"ʥé@ʥĜ ʥ)ʥ& Nʥ*ʥ4Xʥ)ʥ#
"ʥ
;
MDʥ-'ʥ ʥNʥʥ ʥ))ʥ))ʥʥ ʥ ʥ9ʥ)ʥ ʥȐ ȋȌʥ-ʥ¼½ʥ9¼½ʥ-ʥ ʥ ʥ? 9ʥ F ʥ ʥDʥ 9 9ʥpʥ ʥw 9ʥ@ʥ ʥ-ʥw -ʥpʥ+ ʥ$*))N))ʥ Z*&ʥ 9ʥʥ*ʥʥ ʥN)$Nʥ*ʥ'ʥ ʥ$*))N))ʥ^N)ʥ'ʥ ʥ''ʥ~ @ʥ ʥ+ ¡ ʥ1ʥ"-ʥ F ʥ uʥ 99ʥ1ʥ
ʥw ʥss ʥ, 9ʥ@ʥ ʥ
ʥ$*))N))ʥ ʥ ʥ *ʥ*ʥ'ʥ 79ʥpʥ ʥ ʥ ʥ+ʥ"ʥʥ" ʥ"-ʥ'ʥ+ʥ-ʥ F ʥDʥ 99 ʥpʥ
ʥw - ʥʥ sʥ 9ʥ9ʥʥʥʥ*ʥ'ʥ@ʥ ʥ 'ʥ$*))N))ʥ Ȉʥ@* Nʥʥ$*)Nʥpʥ"-ʥʥ
ʥ,ʥ 'ʥpʥ-'#ʥ
;
MDʥ"ʥ ÐlʥpʥʥOʥpʥ ʥ ʥʥʥʥ ʥɲʥʥ ʥ ʥ$ )ʥ Ð Oʥ1ʥʥlʥpʥʥʥʥʥʥ*ʥʥ ʥ+ʥ$ )ʥ ÐlʥOʥ? ʥʥ &ʥ ʥ Dʥ F ʥ DʥOʥ²ʥ.ʥlʥ1ʥN Nbʥ.ʥ>
ʥ ʥbʥ= ʥʥ ʥȑʥ.¼½ʥN Nb¼½ʥʥ V*ʥ)ʥbʥ² ʥ.1ʥ ʥ ʥʥ ʥ)ʥbʥpʥ ʥ.Ʃ1ʥ ʥ ʥʥ##ʥʥ'ʥ@ʥ ʥ ʥ$ ) F ʥ ŌʥOʥ²ʥN N.ʥlʥ1ʥbʥ.ʥ> ʥ ʥbʥ<
ʥʥ ʥȒʥ
.¼½ʥb¼½ʥʥ V*ʥ)ʥ.ʥ² ʥbʥ1ʥʥʥ"ʥ'ʥ+ʥ-ʥ ʥ ʥ ʥ ʥ)ʥ.1ʥ ʥbʥ;+ʥ-ʥ ʥ ʥ ʥ @ʥ ʥ-ʥʥp ʥ ʥ ʥ$ cʥ Zʥʥ &ʥ N*ʥ ʥʥ+ʥʥ ʥʥ ʥ1ʥ"ʥ$ cʥ
;
MDʥ ʥ ʥ NNʥ ʥ<ʥ ʥ )Ãʥʥ*))*ʥ ʥá**ùʥ<ʥ iʥ zʥ*ʥʥʥ*ʥá**Ɖʥ<ʥ ʥʥ)ʥ)ʥʥ*ʥʥ 9zʥ*ʥʥʥ*á**ùʥ<ʥ ʥʥ)ɪʥ)ʥ ʥ*ʥʥ 4zʥ*ʥʥʥ*ʥá**ùʥ<ʥ ʥʥ)ʥ ʥ ʥ &ʥ ʥ zĴʥ1ʥ9zʥ ʥ ʥ<ʥ)ʥ zĴʥ ʥʥ*&ʥ*ʥ ʥ<ʥ )ʥ9zʥ;ʥN)ʥ)ʥN ʥ))ʥ
'ʥ
% ʥ
#6; !6;ʥ $6; ʥ ʥ<ʥc
ʥ!6; ʥʥ&ʥʥ ʥ <ʥʥ#6; ;ʥ ʥ ʥ ʥ%ʥʥʥʥ ʥʥʥ ʥʥ%ʥʥʥʥ ʥ ʥʥʥ ʥ ʥ<ʥʥ$6; ʥʥ&ʥʥ ʥ<ʥ #6; ;ʥ ʥ ʥ ʥʥ ʥʥʥʥ
%ʥ
$7; 1ʥ#6; ʥ ʥ<ʥʥ$6; ʥʥ&ʥʥ ʥ<ʥ ʥ#6;;ʥʥʥ ʥ ʥʥʥ
Zʥʥ ʥʥ ʥ#6; 1ʥ# ʥ5ʥmo ʥ FÙʥ ʥ#; 0ʥ ʥ ʥ3>ʥ0ʥʥʥ#6; 1ʥ ʥY ʥ6; ʥ#6;1ʥʥY ʥ6 } kʥ!4;ʥ#4;ʥ14; 0ʥ#4;ʥ#4;1ʥ ʥY ʥY Ĉʥʥ ʥY Ǽʥ²ʥ #
9O
%.o
HDʥ%'??'ʥ F ;ʥOʥ ʥ,: ʥʥʥʈ#ʥ m^ʥʥʥʥ$ɒʥ ʥʥÙɨʥO5ʥ1ʥ ʥw Ĉʥw ʥw 'ʥw
>ʥw ?ʥw ' mʥʥʥ ʥʥʥO5ʥʥʥʎʥ ;ʥ ʥ SOʥ ʥ S5ʥaʥʥe ʥʥO5ʥʥ ʥ i ʥʥʥOʥʥʥ%ʥʥ ʥ ʥʥʥ5ʥʥ ʥ%ʥʥ"ʥʥ+ʥ HʥǸ TO5ʥʥ?ʥ&ʥ ʥ ʥʥʥOʥ ʥr ŝʥʥʥ ʥ? UOʥʥ? U5 k ;ʥ
ʥEOʥʥ
ʥ&, kʥ,:0ʥ ?+yʥ@ʥʥ ʥɽʥʥyʥʥʥʥ?#ʥ kʥʥ ʥ ʥʥ'ʥ ?ʥ'ʥ ʥʥʥ ʥʥʥyʥʥʥ?ʥ ʥ &ʥ;ʥʥʥ Dʥy 0 'ʥ ʥw 'ʥʥʥw 'ʥw ?ʥʥʥw ?ʥw '#ʥ Ş ʥ ʥʥÕʥʥʥ5ʥʥʥʥȍȎȏʥ ?+yʥhʥ"ʥʥyʥ1ʥ'ʥw 'ʥ Zʥ5ʥ1ʥ ?+y1ʥ + ʥ Oʥ0ʥ + ʥʥOʥʥ5ʥ1ʥ'??'#ʥ
+ʥ
Cey\O~X =O`XtOeSOp C~SeX Efz]PY>PaYuPfTPqAqufPV E>A E;;Q % !"' $! ' !% ' '
Eq @:=? J:6"J=I\} %1&1_ X\}b\m\qa}b|t]c}||c[\| =??: := y\|u\Y }c\m Oyt\}bW} 1 % 6 &&1' 1
Tb\ |\} t] "[cac} cq}\a\y| Wy\ [cc[\[ cq}t }
t |\}|/ }b\ |\} Qo Ytq|c|}cqa Wmm cq}\a\y|\WYbt]
bcYbYWqX\y\uy\|\q}\[W|}b\uyt[Y}t]}
t [cac}cq}\a\y| Wq[ }b\ |\} Ro
bcYb Ytq}Wcq| }b\ t}b\y cq}\a\y| VbcYb t] }b\ |\}| Qo Wq[ R Ytq}Wcq| oty\\m\o\q}|8 Quut|\ %%1 1%1 c|W "+[cac} cq}\a\y|Yb }bW}]ty\WYb #6 " "( }b\ "[cac} cq}\a\y %(%()1 Ytq}Wcq| $o [c|}cqY} uyco\ ]WY}ty| Bcq[ %++,01 Mt}\/ -./1 [\qt}\| Wqcq}\a\y
bt|\ [cac}|Wy\ -1.1/1 )
I\}RX\}b\|\}t]cq}\a\y|}bW}YWqX\
yc}}\qcq}b\]tyo&p1$\o
b\y\pWq[ q Wy\qtqq\aW}c\cq}\a\z|Bty\Woun\ $o & &$Wy\WmmcqR Bcq[}b\|ot]Wmm ut|c}c\cq}\a\y|qt}c{R
-
I\} %1&1 X\ ut|c}c\y\Wm qoX\y| |W}f|`cqa %1&16 Qbt
}bW} c] -1-1
11- Wy\ut|c}c\y\Wm qoX\y||Yb }bW} --1
1-16 }b\q %-11&%-1&0001%-1 &1#
+
Dey\O~X =O`XtOeSOp D~SeX Efz]PY>PaYuPfTPqAqufPV E>A E;;Q % !"' $! ' !% ' ' #!%$! #'
K\} BGx X\ }b\ Wqam\ Xc|\Y}tx t] MBx Tb\q 3BCDx89:;x 3GBDx Ub| BCx"xGBx@ BD"GDx @x CD\BDx I\} HGx7lx Wq[ GDx7m x Ub\q W"lx @x V\mx @x T\Vx Tb| WVx@ TlxV+x@xTmx Tb|V+x xWVx@ Tl Tx mxWq[ b\qY\ VVx W @ xT+x
N
|
V\|tm\}b\a\q\xWmYW|\t]"r[cac}cq}\a\x|
b\x\r# " Tb\x\Wx\ %#+i %#+i' "r[cac}cq}\a\x| Ub\x\Wx\ %#ix ki'x r[cac}cq}\a\x| >tq|c[\x Wmm}b\uxt[Y}|t] uWcx|t]r[cac}cq}\a\x| Ub\}t}WmqoX\x So t]|Yb uxt[Y}||W}c|_\| _
ki(x x A kix
"
%#+i %#+i' %#+i( %#+i+ kix x ki'x " " Ub\|\uxt[Y}|cqYm[\Wmm}b\qoX\x|cq Nx Ub| KNLx A LOx T
t[cac} qoX\x|
bcYbYtq}Wcq }bx\\ [c|}cqY} uxco\ ]WY}tx| Wx\/ $@x "$& %"@x "$) '7%$& ''7" $ ) @x "&) )*@x "$ $ *%7%$) Bztob\z\
\YtqYm[\ }bW} %(16' ]tzg6 11 "(Wq[ %1 c|\c}b\z'tz 1
)#x
E] -1 c| }b\ |oWmm\|} cq}\a\z cqR |Yb }bW} -1 # ot[$ }b\q -11$j IoR Wq[ - $ j 11 R ]tzWmm j JK 1 V\bW\$ c|}b\|oWmm\|}om}cum\t]$ }bW}c|cqR2 c||oWmm\|}qoX\zcqR}bW}c| J 1 ot[$ Wq[ 1 c|}b\|oWmm\|}qoX\zcqR}bW} c| 1 ot[$ Tb| }b\ut|c}c\qoX\z| qt} cqR Wz\ 1% +( Wq[ 11 %( Tb\cz|oc| $$+(11 '"1%( 1 1 1 1 )
1O
-
Tb\m\]bWq[|c[\c| % -$1
1$11% &-$$$11$$$$11 1$$$$ 1% & -$$11$$$11 1$$$ 1% & --11$$11 1$$11%& -11-11 1-11& 1 1 1 1 1 1 # % 1% &11% & 1% & 1%& 1& !% 1& i
Tb\mW|}c|}z\|cqY\X 9L CL cq\wWmc} 1 $$$$11$$$$11 1$$$$1 JK $$$$$ 6 $$$11$$$11 1$$$1 # $* $$$-*"1 # $$11$$11 1$$1 JK $$$$$ *" 1 $11$11 1$1 JK -*$$ $$ 1 /O
/O
/O
Mt}\ BtzWuztt]t]}b\a\q\zWmYW|\ |\\Q\qctzP%
1
Bdx[N}W
cnlf*D^xnD::P
:PM:/ >D::r
#"$&!&)
"'+&-- LN %)'(#"'
"(&-,#)&-"'+&'-#"-(-"'+&-'(-$*#&- (- !) ($ - #- %)'(#"'- "(&- ,#)&- "'+&- #"- (- "'+&- '(- ,'"- (- ) -#"(""-(- ((&- - V( Y) [ #&- - #&&'$#""-(# (-#&&(-"'+&#&- (- #(&- '#&(- %)'(#"'- +&(- ,#)&- "'+&- #"- (- "'+&- '(- "'-(-$$$&(-) - #+-,#)&-"'+&#- '($'-&-"-(#-)'(-,#)&- "'+&' - %)'(#"- &&'- = !&
#- ) (#&'- &- #+-
)))) ) )) U )))
ʥ
_ts{tmXqzsjmam~szw~
#ʥ
k$$ʥ ʥp ʥ ʥ$ ʥ ʥ ʥ ʥ9ʥ ʥʥ$ʥʥʥ$ ʥp Uʥʥ ʥʥ ʥ ʥ9ʥʥʥnʥʥ ;ʥʥ ʥʥʥʥ$ ʥ ʥ ʥʥ ʥʥ ʥʥ ʥ ʥʥ'ʥïʥ ʥʥ ʥÝʥ ʥʥ ʥ9ʥʥ "ʥnʥ$&;țʥ %Hʥ ʥ %aʥ ʥ %Fʥ ʥ %Yʥ ʥ jʥ Ue;ʥ ;ʥ
iʥ6ʥ=% '(!% ?P !% ?ʥʥ!% 'ʥ!ʥ ʥʥ! ʥ ʥ ʥʥAʥ ʥ$ʥ& ʥʥ6ʥ %Hʥ Ƭ ?"ʥ %aʥ P ?ʥ %Fʥ P ?ʥ %Yʥ P ?ʥ jʥ P ? ʥ
ʥ
UʥÝʥ ʥʥ] ʥR ʥ ;ʥʥʥʥ& ʥ%ʥ 7ʥÕʥʥ $ ;ʥ ʥʥ< ʥ!/ʥ( JK! ʥ ʥRʥ>ʥ ʥ ʥ %Hʥ ʥ +ʥ %aʥ ʥ
"ʥ %Fʥ ʥ
"ʥ %Yʥ ʥ ʥ jʥ ʥ ʥ
Uʥ!ʥ 6ʥ ʥ ʥʥʥʥ n!mʥ!ʥʥ 6ʥ ʥ> ʥ ʥ 4J 6ʥʥ! 'ʥ Õʥʥ & ʥ !ʥʥ6ʥ %Hʥ (ʥ %aʥ P ʥ %Fʥ ʥ %Yʥ ʥ jʥ ʥ .O
ʥ
#
Uʥ ʥ) ʥ 342ʥ)ʥ 0 ʤʥ ʥʥ3ʥ1ʥ 3 A)ʥʥg ʥʥʥ4# ʥ
ʥ "ʥ
"ʥ %Hʥ ąą "ʥ "ʥ aʥ " "ʥ %Fʥ " "ʥ " %Yʥ "ʥ "ʥ " # ą (Ƽ "ʥ "ʥ
"
@ʥ ʥʥ ʥ) ʥ 34ʥSʥʥɸɦ#ʥíSʥ=2ʥ³ʥ ʥX ʥʥSʥ 32ʥ34ʥ 4 ʥÇ)g;2ʥ ʥʥ)ʥʥ×ʥ#ʥUʥ =1ʥʥn2ʥ=31ʥʥn2ʥ ʥ ʥʥó) ʥ 3¥ʥ)ʥ< ʥʥʥ ʥʥ< ) ʥ=3¥X2ʥ×ʥʥ ũ ʥ ) ʥ Ƞ4ʥ)ʥ>#ʥ %Hʥ
ʥ aʥ ʥ %Fʥ ʥ %Yʥ ʥ
ʥ \
=ʥ '#ʥ
A)ʥʥg ʥʥ %Hʥ ƭ ʥ Eʥ aʥ ʥ( ʥ ʥ E
ʥ( ʥ W ʥ E %Yʥ ( ʥ Eʥ
ʥ( ʥ # ʥ E
ʥ ʥ ##ʥʗ ,ʥ ,ʥ ,ʥ
E EEʥ EEEʥ E E Eʥ
!
"
#
$
%
!
&
%
ʥ
+
@ʥ ʥʥ&$ʥʥ ʥʥ+ʥAʥʥʥʥ ;ʥʥʥ ʥʥʥ ʥʥ ʥʥʥ ʥʥ$ʥʥʥ&$ʥ ʥ ʥ&$ʥ ʥ^ ;ʥʥʥ ʥʥʥʥʥʥʥ &$ʥ cʥʥʥʥʥʥ ʥʥʥʥʥʥʥ &$ʥ ʥʥʥğ Hʥ ʥ %aʥ Ƕʥ %Fʥ %Yʥ
jʥ "
?
Ycʥʥʥʥ ʥó ʥ%ʥ ʥ ʥ ʥʥ ʥ?ʒ ʥʥ ʥ ʥ%ʥʥ ʥ$&ʥʥ ʥʥ ʥ ʥʥʥ ʥʥ ʥ ʥ Hʥ ʥ %aʥ ʥ %Fʥ ʥ %Yʥ "ʥ jʥ 'ʥ
i 94=ʥʥ ʥ< ʥcʥʥ ʥcʥʥ ʥ 4ʥ ʥ T ʥʥʥ$$ ʥʥ=ʥʥ 9ʥ ʥʥʥʥeʥʥUʥ = 0 =4ʥ ʥʥ ʥʥ< Ɇ ʥ 94=ʥʥ ʥ>ʥeʥʥʥʥ=Tʥʥn ŕʥ
Hʥ %aʥ %Fʥ %Yʥ
ǎʥ Ēʥ Ǐʥ ǐʥ "ʥ
ʥ
bqz|am~szw~
#ʥ Aʥʥʥ $&ʥ&ʥʥ +\ʥijʥʥw +ʥw ?ʥ,ʥ >
ʥ k$$ʥ ʥ.ʥRʥ ʥ` ʥ ʥʥ ʥ ʥ #ʥAʥʥ& ʥʥ
ʥ ʥ ʥ
ʥ
ʥ,ʥ
ʥʥw}
ʥʥɃ}ŧ
;7;
;8;
ʥ Aʥʥ ʥʥ Eʥw ʥʥ Eʥy ʥʥEʥw ʥʥ ʥ +"Eʥw +"ʥʥ &ʥ;ʥ ?ʥ °ʥ °
#ʥ Aʥʥ& ʥʥ , ǕȪE ( ɯǞE ʚ
ʥ ʥ.:ʥ ʥʥ
5Aʥ ʥ& ʥɊʥ.ʥ A
ʥ
_
ā.(ʥ [W
"ʥ UʥʥÕʥʥ 34ʥʥ ʥ ʥ ʥ=ʥʥ ʥ$ʥʥʥ34ʥí Tʥʥʥ ʥ 3ʥʥ ʥ=Tʥʥʥ ʥʥʥ.&; ʥ$ʥXʥʥʥʥ 4ʥ ʥ ʥ=Xʥʥʥ ʥɾɮʥʥȯ =4 #ʥAʥʥ& ʥʥ Tʥ#ʥ3=ʥʥ4Xʥ # T3ʥ =4ʥ X ʥ ʥ
'#
Aʥʥ& ʥ %CǹƓʥ( %Cúʥ(ǩ%CƔʥ( %CCúʥ( %C úʥ %C ĸ ʥ
+
O
AʥʥʥʥCʥʥÞ.ʥRʥʥ ʥ ʥ ʥʥ6 ʥʥ??ʥ ʥʥ .Rʥ. , Rʥ ʥʥ ʥ$ʥ ' "ʥ
?#
ʥ!ʥʥ ʥ ʥʥʥ ʥ!/ʥ( !ʥ ʥAĚʥʥ& ʥʥ! Ëʥ -O
X
#ʥ ĐʥʥŪʥʥ 34ʥʥ ʥ ʥ 3ʥ0ʥ ʥnʥ ʥ34ʥ ʥn#ʥíʥ=ʥ
ʥTʥʥʥʥ 4ʥ ʥ$ʥXʥʥʥ34ʥʥ ʥTȢʥʥ$ ʥʥ 3ʥ ʥ=Xʥ ʥ$ ʥʥT3#ʥ&ʥ ʥ3Tʥʥ ʥ ʥʥʥȰ34ʥ ʥ ʥ Ȟ0ʥ ʥnʥÖʥʥʥʥ4=ʥʥnʥ Eo n
ʥ ş 0ʥ ʥ ʥʥʥ7ʥ"7ʥ"ʥYɈʥʥʥʥʥ$ʥ%!ʥ6ʥʥ ʥ ʥ! 6ʥPşʥ!ʥʥʥ 6ʥ ʥ
&ʥ ʥ_un> _u ʥv? ʥ ʥʥ###ʥĵʥ ʥDx@_o@ ʥ Ūʥʥ&
ʥv_uX_u
ʥ
Đʥʥeʥʥ 34ʥʥ ʥó ʥ 3ʥ0ʥʥnʥ34ʥ? ʥnʥ 4 0ʥ ʥnʥí vʥ ʥʥʥʥ 3ʥ 4 $&;ʥʥ ʥ
ʥʥvʥ>] &ʥ ʥʥ ʥ$ʥʥʥvʥʥyʥnʥeʥ
ʥʥ34ʥ ʥ & ʥʥyʥ ʥ
3ʥ
1;
'ʥ
Uʥ!ʥ6ʥ ʥʥ ʥ ʥʥʥ ʥ!ʥ,ʥ6ʥ,ʥʥ ??ʥ ʥco ,ʥ6ʥʥ!ʥ? ʥe ʥ ʥ$ʥ& ʥʥ #ʥ
AʥʥʥsP ʥ ;ʥ<ʥʥ;ʥ^ʥsûʥ ʥ^ʥ ʥûʥʥ ʥʥsûʥ ʥʥAʥ^ $Ĝʥ
ʥʥʥ ʥ<ʥ
ʥ ʥ
ʥ ʥ
"ʥ Uʥ
ʥ
(ʥʥ ţʥʥ ţʥʥʥ
#O
e !
'
Acʥʥʥʥ$&ʥʥ!ʥʥ!ʥ ?ʥʥ hʥ"ʥ ʠ
+#
ʥ5ʥʥʥ$&ʥʥʥ ʥƕ
ʥ
ʥ
ʥ ʥ { {ʖʥ N Nǟ Ǡʥ ȱ łȬʥ Aʥʥ& ʥ 5#ʥ
?
UʥʥÕʥʥ 34=ʥʥ ʥ ʥTʥʥʥ$ʥʥ =ʥ ʥXʥʥʥ $ʥʥ4³ʥUʥʥ ʥʥ ʥ3=Xʥɚʥ ʥ_ʥeʥʥ ʥʥ 34=ʥʥ_ž
Uʥ ʥʥʥʥ"Ʊʥ$&ʥ¸ʥʥʥ+ ++
++ʥ&;ʥʥʥʥʥ $$ ʥ ʥ ʥʥAʥʥʥʥ "ʥ$&ʥ
ʥ!ʥ 6ʥʥ$&ʥʥʥ ʥ '!ʥ,ʥ6Œʥ?ʥAʥʥ $ʥ& ʥʥc
M
/(
+ʥ
ʥ
: ¢ ʥ ?
Yʥʥ×ʥʥv - ʥʥ^$ ʥʥ% ʥv* v v-
#
Aʥn ʥ ʥʥʥʥʥ]uI n ʥ ʥʥbu ʥʥ ɿʥĝ &ʥʥ.fʥ ʥ]o Aʥʥ ^ʥ& ʥʥbu ʥr ʥ&ʥ $&ʥʥ
b
Za@
n
Ǘʥ ĕʥʥʥ ʥ ʥʥʥʥ"ʥ'ʥ+ʥʥ ʥʥ+Eʥ0ʥ ʥ+ʥʥʥ ʥʥʥʥ ʥ ʥ#ʥAʥ ʥ?o[o?o ʥʥ]e ʥʥyʥ ʥʥʥʥ ʥ ʥʥ ʥDʥ
"'+ "+' '"+ʥʥ +'" ȕʥ ʥʥ ] "'+ʥ ] "+'ʥ ʥ ]#" I+'" ʥAʥ] - ] +| -O
-O
-
*;
Zʥ :`6 iʥvʥ ʥ$&ʥʥ ʥʥ ]I :¯Qʥv5 ʥbʥI ¯Qʥ ʥʥ ʥʥʥ ʥʥ< ʥʥ` AʥʥB ʥ$ʥ& ʥ ./(R / žʥ
?ʥ
Qmwf`zy{dN`~gds`~ma`qQyamd~ bswpe{z}m_eqmvesjet`t
v{sek b_` 8o bmwsz}bmjszwbztszw
#
HDʥF Uʥʥ$ ʥp ʥ ʥʥʥ ʥ'ʥnʥʥ ʥ ʥ ʥ ʥʥʥ "ʥnʥʥ ʥ3
UʥZoʥ ʥʥʥʥ$ ʥp ʥʥɺ ʥʥZo ʥ ʥʥ'ʥnʥ ʥʥ ʥʥZo ʥ3ʥʥ "ʥn2ʥʥ Zo ʥʥ ʥʥʥʥ F ;2ʥʥ ʥʥʥʥʥ$ ʥ ʥ ʥ ʥʥʥ2ʥ ʥʥ ʥʥeʥ &#ʥ
MDʥH ʥ & 6: Æ ' /ʥ( !/ ?/ʥ( !/ʥ :ʥ! ± ā ' /ʥ ?/ !/ʥ ' /ʥ{ ?/ :ʥ! ±ʥ( "!/ʥ / :ʥ!ƅʥ 4/:4/4 4/9 Zʥʥ ʥ$ʥ& ʥ6ʥʥ / ( /ʥ:ʥʥ( "+ʥ:ʥ( ?"ʥ
#ʥ
HDʥHʥ @ʥɛ ʥʥʥ< ʥʥg0 R#ʥ @ʥʥ< ʥɕ ʥ ʥʥʥ ʥ;ʥʥgGRʥ @ʥ ʥ$ ʥʥʥeʥʥʥʥ ʥ ʥʥ $ʥʥ ʥg+ Rʥʥ ʥʥhʥ\ Rʥ@ ʥ
ʥgR#ʥHʥʥ ʥʥʥ ʥ$ ʥʥ
#
.;
ʥʥ ʥʥ<ʥ$ ;ʥʥ\ + ʥ
ʥ O
!#
ʥ !#
,
MDʥ%Y Uʥ!ʥ?oʥʥ!ʥ0ʥP!ʥ ʥʥ ʥʥ!ʥʥ!ʥʥ6ʥ ʥ ʥ6ʥ:ʥ @ʥ6ʥ ʥ ʥ $&ʥʥŴÏ(6ʥ !ʥ0ʥ'ʥ&ʥ!ʥWʥ'ʥʥʥ ʥó ʥʥ!Œʥʥ @ʥʥʥ &ʥvM ʥF<;ʥ!ʥʥ!ʥʥ6ʥWʥ ʥ&ʥʥ<ȿʥ !ʥʥ6ʥ0ʥ ʥ Ɯ ʥ Uʥ6ʥGʥʥŴÏʥP 6ʥ,ʥ!ʥ0ʥ'ʥ&ʥ!ʥ 2 o kʥ!ʥWʥ'ʥʥ!ʥʥ!ʥ,ʥ6ʥWʥ ʥʥ ʥ6ʥ:ʥ( ʥʥ ʥGʥZʥʥʥ &ʥ6ʥʥʥ ʥʥ ʥ ʥʥ< ʥȸÏʥP 6ʥ,ʥ!ʥ0ʥ'ʥ ʥ !( 6 0ʥ'ʥ % ʥ k&ʥ< ʥ% ʥ ʥ% ʥ&ʥ!ʥ:ʥ ʥ6ʥ:( , L@ʥ!ʥ,ʥ6ʥ0ʥ ʥ ʥ n
n
M
Ŏʥ%F
ʥ ʥ Ɨʥ ʥ ʥ 9ʥŔʥƳ ʥ Ɩʥ ʥ ʥ ʥŔʥ¢ʥ å$ʥåʥʥ ʥʥ ,L ʥʥ9ʥ ʥ å$ʥåʥʥ ʥÄŵʥ ʥ ,Lʥ(¢ʥ ʥ ȶÄɤɥʥ :ʥƲ ʥ ʥ ʥ
ʥ
ʥ ʥʥʥ9ʥʥ +ʥʥ Uʥʥ :(ǒʥʥ% ʥ 3:ʥc
3ʥʥ 3ʥ:ʥj. ^ ʥhʥ ʥ ʥ ʥ ęʥʥʥ$ʥʥʥ ʥʥ3 ʥ ʥ +öʥʥ ʥɴ ʥ@ʥʥʥ &ʥ ʥ : ǓʥF<;ʥʥ4ʥ0ʥ +ʥĂ% ʥ9ʥʥ & ʥ ĭʥ ʥ4ʥ:ʥ(%ʥ ʥʥ9ʥ:ʥ(ʥ ʥʥ9ʥijʥʥ ʥʥ3ʥ (¢,L ʥ¢ʥ ʥ: "ʥ ʥ ʥ ʥ ʥ "ʥ m
3O
-
"ʥ
K
L
MDʥjʥ k
ʥ ʥʥ ʥ 9Tʥʥ< ʥʥ
ʥʥ< ʥ=9TXʥʥʥ ʥ
ʥʥšT ʥ0ʥ ʥʥšTXʥ@ʥ $ʥ ʥ=Tʥʥ$ ʥʥ X#ʥ 4Tʥ Ĺ: =ʥ ʥ ʥ @ƴ: ¢ʥkÄŵ 43ʥ 3ʥ +ʥ
Ũ ūʥŜTêʥ êTʥ ʥ ʥ
Ũ ūʥŜ9êʥ ê9ʥ
ʥʥT4ʥ0ʥiw , ʥ_ʛʥ +ʥ -O
ʥ
1;
=
3ʥ
'
HŎʥ%F Aʥũʥ ʥ%5¡ Eʥ%5ʥ( Eʥ5Eʥ%5( Eȼ ʥ%5ʥP ʥ5ʥ %5ʥ( Ⱦʥʥ5 /%5( E# @ʥʥ&ʥʥ ʥʥʥ # 5( ʥ # ʥ # 5ʥ : : *5ʥ 5ʥP üʥ#ʥ *5ʥ5ʥ i ( %ʥ5ʥ( üʥ i ( üʥ ʥ *5ʥ5ʥ ʥ ʥ ƶĺ ʥ : * " #: 5Eʥ * %5ʥ( Eʥ 5Eʥ
ʥ ʥ : ŸʥĺP kkl ǘ ʥʥʥʥ$Ÿʥɩ2ʥʥʥ ʥʥ : Eʥ Eʥ c\
d]
e^
+
M
4M
f
f\
HDʥYʥ @ɖʥ :ʥ'ʥ ;ʥ $Ěʥʥ ʥʥğʥ ʥ
ʥʥʥ&$ʥkʥ ʥʥ ʥʥʥʥ ʥ+ʥ "2ʥʥ ʥ+ʥ ʥʥʥ ʥʥʥʥʥʥ&$ʥ ʥʥʥʥʥ< ʥʥ +ʥ ʥʥ ʥ< ʥʥʥʥʥʥʥʥʥ&$ʥ ʥ ʥ2ʥ ;2ʥ%+ʥ'ʥ ʥ ʥƝ+ʥ"2ʥ2ʥ ʥ%+ʥ2ʥ2ʥ ʥ%+ʥ2ʥ2ʥ ʥ%'ʥ"2ʥ2ʥ %'ʥ"2ʥ2ʥ ʥ%'ʥ2ʥ2ʥ ʥ ʥ"2ʥ2ʥ2ʥ#ʥZʥĚʥʥ ʥʥ<Űʥ ; '(+ʥ:ʥ # ʥ +O
?
HDʥ ʥʥ ʥʥʥ ʥ ʥʥ!Ŷ2ʥ|!Ŷ2ʥʥʥ
ʥʥ!ɰʥ@ʥʥ &ʥ!ʥʥ ʥ!ʥ +ʥ ʥʥʥ!ʥ0./!ʥʥ?ʥAʥʥ < ʥ!ʥ00!2ʥʥ ʥ!ʥ!ʥʥ ʥ!ʥ0'!ʥ ʥʥʥʥ ʥeʥ< ʥ ʥ +ńŅʥ!ńŅʥ"ʥkʥ! ʥ ʥʥ ʥ!ʥʥ?ʥʥ 'ʥ ʥ!ʥ "ʥ 'ʥ +ʥ ?#ʥZʥʥ ʥʥ ʥ ʥʥ ʥ $&;ʥ% "Žʥ'"ĥ2ʥ'+ ʥ% Šʥ' ʓʥ+Ğʥ% +Žʥ"+ĥ2ʥ+ĥʥ ʥ % ?ʥ"ʥʥ+Šʥ n
+O
;
ʥ
MDʥa
ŕʥ
ʥÉʥʥʥ ʥʥʥʥʥʥ k = W =4 ʥ©¾=4 : ? ýʥ ©¾4=: ý#ʥ@ʥ©¾9=: Ƙ#ʥHʥ ©=T9ʥ:ʥ?ýʥʥ$ʥ ʥ=T1ʥ9T#ʥʥ!ʥW =T W 9ʥ T#ʥkʥ94 ȖʥT=ʥ ʥ
ʥʥ< ʥ 94=ʥB ʥʥȲT=ʥʥ ʥʥȳ9=ʥʥ ʥʥȴ4= ʥ &ʥ ( /ʥ ( { 94 { !: (Ļ% 994!ʥ ( Ļ{ % 9(!! , (¢!ʥ¢ 1ʥ ʥ ʥ ʥ ʥ ēʥʥʥ ʥ ʥt = ʥʥ$ɷ ʥʥ 4ʥ ʥ
ʥ < 94=ʥ ? ƙʥ94ʥ, ʥʥ4= ʥ &ʥ ʥ:ʥ[ 9ʥ 94ʥZ ʥ 4ʥ{ʥt = :ʥ ʥ[{ʥ 9ʥ 94ʥƨʥÉ/ % ʥ ʥ ʥ ʥ mʥ< ʥ% ʥ ʥ 9 /ʥ,ʥ94 /ʥ:ʥ 4 /ʥ:ʥÉ/ʥ$;ʥ ʥ % 994 /ʥçÉ/ʥ, { 9 {ȟ4: É/ʥ% ( É/: ?" #ʥ @ʥ 9ʥʥ94ʥB Ē #ʥZʥʥ< ʥ (2ʥʥ ʥ!ʥA Ē # JO
((
IO
JO
MDʥ ? ʥ ʥ_? +ʥ@ʥ % + ʼnʥʥ 6 w +ʥw ?ʥ(kB _56_ _,ʥ( ,ʥ(k
A _5 6_0 6_ (k ? _ ĉʥ? ? ĉʥ?l0{ʥ ĉʜ Zʥʥʥʥ$&ʥ&ʥʥ (1ʥ %"ʥ,ʥ( ?? ʥ
66
%(
MDʥ *
%L
%L
%LLwǯ 7 ; %LL
%Lʥ %L
%L %L
%L
;
ʥ}ʥ ʥª /.ʥ}ʥ ʥ}w ʥ /ªw}ʥʥ ʥ ªʥ ʥ ./Rʥʥ :L wʏʥR Rʥ %L ʥ %L ʥ %L w}ʥþ.Rªÿʥ Ʉ}ʥ }wʥ : ʥwʥ:)L;LʥwʥR /ªʥʥwʥª /ʥ: :ʥǪwʥ:;<+L:ʥ*L
%*
MDʥ (##0 ē&ʥ ʥ@L @L @L @LL% %L @LL% @L@ %L %LL(L (LL*L *LL L(0.L (0.L (L%LL* (LL, L!*Lʥʥʥ(0/ (0.
n
{
{
n
{
JJJO
{
+O
(0/ %L kʥ(##1L (0/L / L(0/ %L %L (##0Lʥ(##1LUʥʥ ʥʥ ʥʥ(##0L *O
n
%,
l
l
MDʥ ʥX H (-LʥGE+Lʥ(-L%#87&+KL@ ! °ʥ:ʥ(DL%L#LL(-L G*L (-)L %##-L9&+L(DLʥG>+LL(DL G>+L ʥ&ʥ!+L3-#L%-H Lʥ!(! /ʥʥ !ʥʥ%#L L #L@ʥ< ʥ ʥ; ʥ ʥʥ!ʥ -L -O
*O
%-
MDʥ %#(, :4L%Lʥ$% &ʥ(:L %/ʥ:ʥ(##1 Lɘʥ$eʥʥ :/ :4L-#( L
(L :+L -#*: DAAIBL:ʥ::/ʥ : -#(:)L : DAAIBL :ʥ:: /ʥ :-#(:)L : -#((FL #Lʥ# L ʥ('$L:ʥ%#(,L
H
*,L
"#
HDʥ
kʥ=³ʥʥʥ ʥʥʥPHo ʥ &ʥ Tʥ:ʥCoʥk ;7ʥ ³3ʥ 3=ʥ Do Co Tʥ 3 3=ʥ $); =Xʥʥʥ ʥʥʥ.$; : ĹʥZʥ :ʥ 4Xʥ =4ʥ T3ʥ =4ʥ X ʥ
'
HDʥ + Aʥʥ ʥʥ ʥ3ʥ:ʥƆ7ʥ ʥ ʥ: %ʥ ,ʥ3:ʥ ʥ ,ʥ 3ʥ 7ʥ ʥ
( 3ʥ
ʥ¬ ʥ ( ʥ ʥ3ʥ:ʥ ʥ ʥ ʥ3ʥ#ʥF<;7ʥ ( 3 3ʥ:ʥ 3ʥ: # % (%3( : ʥ ( 3ʥ 3ʥ Zʥ % ƇʥƷ % ʥ( % ʥ( Ĝ% ʥƸ % Ğʥ( % ʥ( ʥ :ʥ% ƈʥ¡ % öʥ(% ʥ( % Ğʥ( % öʥ( % öʥ( ʥ : +ʥ
+
HDʥ " mʥ ʥ.Rʥ, . ,ʥRʥW %. ʥ %Rʥ,ʥ P ʥ@ʥ.Rʥ. , ʥRʥʥ ʥ$ʥ 'ʥ
ʥ;ʥʥ%.ʥ %Rʥ( ʥ @o ʥ%ʥ'# 0ʥÞ 7ʥ 7ʥ7ʥʥ2ʥ--2ʥ 7ʥ ʥ ů ʥ:ʥJHLJo D ʥHL@oÅʥ%ʥ'õʥÅ1ʥ7ʥ 7 7ʥ###ʥ7ʥ"#ʥLoʥ ;ʥʥ&Ãʥ ʥʥ ;ʥ HLJo ɜ ʥ ʥeo Jo ǫ7ʥʥÅʥ ʥ0j./"7 co@o ʥ%ʥ'ʥʥ ʥ;ʥʥÅʥ:ʥjKo ʐ 7ʥ"õ7ʥʥ#: Wʥ ʥ ʥjWʥ7ʥʥ#: Wʥ6 ʥj0ʥ @ʥʥʥŬʥ ʥ F ʥ#.ʥ ʥ7ʥRʥ ʥ ʥJo ůʥʥÅʥP J% L"õʥʥ @ʥ.7ʥRʥJo ŮʥʥÅʥJo JA LõʥHʥŲ ɱʥ¿ʥ:ʥŲ ɻʥ¿ʥI 7ʥʥʥʥʥʥÞ.7ʥR ʥ :ʥ +
&%1M
F ʥ Dʥ. , ʥ ʥ ʥRʥ,ʥ ʥ ʥ ʥʥ /ʥ ʥ Ëʥ $&;7ʥʥʥʥ ʥʥ @ʥ.ʥ ʥRʥ ʥ ʥʥ ¯ʥ ʥ °ʥ$&;7ʥʥʥʥʥ ʥʥ kʥ¿ ¯æʥ:ʥ ʥ ʥ¿ °æʥ 7ʥʥʥʥʥʥÞ.7ʥRʥʥ ʥ ʥ1ʥ ʥ F ʥDʥ. ʥ ʥ ʥRʥ,ʥ ʥ ʥ ʥʥ °ʥ ʥ Ǻʥ$&;7ʥʥʥʥʥ ʥʥ @ʥ.ʥ ʥR ʥ ʥʥ /ʥ ʥ Ëʥ $&;7ʥʥʥʥʥ ʥ mʥ ʥ¿ /æʥ:ʥ ʥ ʥ¿ Ë æ ʥ@ʥʥʥʥʥʥÞ.7ʥRʥʥ ʥ
ʥ1ʥ #ʥ Zʥʥ ʥʥ + ʥ,ʥ ʥ ʥ " ʥ
-O
-O
*O
ʥ
4E' _©¶§FÏ=EB6B xϸ5Ï4?¸ÏÏ4WÏ Ï2Ï¸Ï ÏYWÏ4?'Ïz§£§ ¸>ÏÏ X¸Æ¨Ï=¸5ÏÏ A Ï 4Ï5ÏCA Ï(Ï 4Ï !;=ϸÈÏ WϸÇÏ W4?>Ï=¹4?5Ï6Y=EB6B'Ï 62' _©¶§FÏ6= W
Eo
xÏ ~Ï ©«©Ï ¶Ï¶Ï ~´Ï ZlÏ Ï4ÏFÏ='Ïi±§¬§£§4ÏHÏÏ© © Ï « ~«Ï ©«©Ï =' Ï Ï o«Ï Zϸ½Ï~Ï z϶ÏA'Ï~´Ï ¸Ï ¼Ï[«Ï4;'?Ï ¦±~«£©Ï ©Ï¸Ï B'¼?'ÏÏ~¼ZÏ m«Ï££¶©Ï ~«Ï ~Ï==¼Ï¸ÏZÏ?¼'6='ÏÏx£´Ï«Ï 64' b©¶§H ÏCE==2Ï zϱ§Ï £Ï©±Ï£§§Ï«§¥©Ï¸Ï¼Ï ¶«Ï¸ ¼Ï©Ï ÉW62C2Ï zϱ§Ï£Ï©±Ï£?§§Ï«§¥©Ï¸Ï¼Ï ¶«Ï¸Ï ¼Ï©Ï 6Ï ÏÊ ÏWÏCB;A2' lÏ«Ï~©¶§Ï©Ï62C2ÏÏCB;A2Ï CE==2)Ï #'; ');o! #; !';
-!#$; %);o'#; %'; ;
!'; ;'$;;!#; ;#$; %); -')$ ;
%'; ;%$;
;
$%;
z
z
Y
,O
;AÏ
,O
4
66* _©¶§FÏ4EE24 4Ï WÏ 'Ïz§£§ i©«Ï£«Ï«~«
M 4Ï 4Ï W; WÏ4EEÏ6Ï622ÏW4EE22' 9 3M 4Ï 44EE22ÏWÏ4EE24' ,,
,,
mF [m[m
[m[m
Y
mG
6;' _©¶§FÏAÏ ` ?5Ï425Ï4;5ÏWÏ 44 WÏ {Ï~´Ï£©Ï` WÏ`c5Ï6``Ïce5ce5Ï `eÏ 6?42Ï 6?Ï o«Ï` vÏ ¸ÏÏ~Ï`ÏwÏ ¼Ï'ÏxÏ~§~Ï£ÏpvwÏWÏ"6ϸ¼Ï©Ï`Ï~Ï~§~Ï£ pceÏWÏ!6Ï`Ï c`Ï ÏeÏ©Ï`ÏY#Ï6ÏÏ?42ÏÏ©Ï`϶ϣ«~Ï?2ÏYÏ#=Ï«~«Ï©Ï¸¼ÏXÏ6?Ï ' 6Ï lÏ vw5ÏWϸ5Ïϼ5 6¸¼£©`ÏY¸Ï5ÏÏ 6? A6?ÏÏ44ÏNOPÏ6ϸ1 6?ÏÏ44ÏWÏ6?ÏÏ44ÏX;ÏA' 7Ï Wϸ Ï=¸Ï =¸Ï5Ï 5Ï d£©¦±«½ÏvwÏ:A϶«Ï«Ï¦±~«½Ï~««~϶ϸÏXϼÏXÏ *O
+O
co
p
MO
Ï ?Ï 6=' _©¶§I x¸ÏϼÏ$MEÏ¾Ï 6=¾¸ÏÏ 6=ϾEÏ¾Ï ¾5ÏE¾6='Ïr£¶ÏY£«Ï «~«Ï¸Ï~¼Ï~§Ï§££«©Ï£Ï«Ï¦±~§~«Ï¦±~«£Ï5ÏϾEÏϾ5E¾ÏÏ6=Ï 2+ _©Ï¸Ï~¼Ï~§Ï§~϶Ï~´Ï¾ÏE5Ï=¾5ÏE¾Ï6=NOPÏ2϶ϩ¥Ï«£Ï ¾5Ï ¼ 6 Ï ¾ Ï ?, Ï A¾Ï Ï ? QR Ï 2 'Ïx£´Ï « Ï ¦±~«½Ï ½ ©Ï4Ï S Ï T ¾ ST Ï ?' Ï {Ï ¸ Ï lÏ«Ï~§©«Ï¥£©©Ï´~±Ï£Ï¾Ï©Ï?-Ï co
*O
*O
+O
+O
;BÏ
+O
*O
6@' i§©®Ï _©¶
§H¥±®ÏÏ;®
Ï©ºÏ4Ï @BÏ ©Ï£
Ï©
¦±
'Ïz
Ï®
§
Ï~§
Ï!M~¥©Ï
£§
Ï®
Ï©®Ï4Ï
®¶
Ï ® ¶£Ï ~ ~
®Ï 4 Ï ©~Ï ~
§Ï ®
Ï ~©®Ï 4 ' Ï i £§Ï
~Ï © ±Ï ~¥ Ï ¶
Ï ~Ï ¥ ±®Ï ~ Ï ©
Ï2Ï£§Ï
£±
Ï t©Ï® ~®Ït©©Ï22'Ï®
Ï Ï ®
§
Ï~§
Ï
º~®½ÏA 6Ï©
Ït©Ï¶
§
Ï mÏ®
§
Ï ~ §
Ï º~®½Ï Ï £±
Ï 2Ï4Ï6Ï !8 !M ;'Ïz
§
£§
Ï ®
ϱ
§Ï£Ï©±Ï~§½Ï ©
¦±
©Ï ¶®Ï
º~®½ÏÏ£±
Ï <Ï Ï !8 !M Ï t©Ï©Ï .Ï AÏ6¿Ï.Ï2M'Ïl
Ï®
Ï~©¶
§©Ï©Ï8 'Ï AÏ6¿Ï2 ;@B'Ï 6A'Ï _©¶
§JÏE@Ï £©ttu £©422ÁÏ £©5Ï@2©5Ï@2ÁÏ 4=©6@3Ï£©6@ÁÏ£©@2ÁÏ 46©@2ÁÏ£©@|Ï £©@2ÁÏ©@2Á5Ï £©@2ÁÏ Ï©@2ÃÏ 4ÏÏ®~@2ÁÏ £©@2ÁÏ ©@2ÁÏ 4®~@2ÁÏ 0ʥ ®~=@ÁÏÏ®~@2ÁÏ ®~E}'Ï 4®~=@ÁÏ®~@2ÁÏ l
»Ï E@'Ï 6B' _©¶
§HÏ66;Ï £ MKMAÏ£ ËÏ0M /Ï » £>M; 1M UÏAÏÏ£<ÏEÏ£<Ï» » £Ï<ÏEÏ0M £Ï»5Ï <Ï £Ï;Ï <Ï UA£<E£<»'Ï £<Ï»Ï £<ÏEÏ 4ÏUÏD²Ï¶Ï©Ï
¦±´~
®Ï®£ o
®Ï²Ï £<Ï»0Ïz
Ï®
Ï
¦±~®½Ï
£
©Ï6²Ï ²6Ï M ²5ÏD²Ï Ï4@Ï 2Ï'Ï £Ï´Ï 'Ï
¦±~Ïn®½Ï´
©Ï²ÏVÏ6Ï£§Ï;ÏVϲÏVÏ@ÏÏ . Ï . Ï ® Ï
Ï Ï ®Ï~®Ï n © Ï ² 6Ï £<Ï»ÏUÏ6Ï£§Ï;ÏUÏ£<Ï»ÏUÏ@'Ïm®Ï££¶©Ï®~®Ï2ÏVÏ»ÏVÏEÏ£§Ï6BÏVÏ»ÏVÏ6=;'Ïl
Ï®
§
Ï ~§
Ï66;Ï©±Ï®
§©'Ï #:
+O
O
f
cET
'O
+O
C
B
+O
: :
:
;DÏ
(O
6C' a©¶§F Ï/2L ®~® i§©®Ï£©§µÏ %L
W
z§£§Ï
(2 L
C
H( H L %L %L %L - 5
M
2
a©¶§KÏ2. O O O O O O O O O O O O O O O O
o®ÏyÏÏ®Ï~§~ϣϧ®~Ï`%Lcfg'Ïz϶Ï~´Ï%L %L %L ~§~Ï£ÏfgjÏWÏ!(L¹Ï~§~Ï£ÏfghÏWÏ!(L¹!yÏ WÏ"y'Ï ʥ +ʥ %L %L %L %L Ï sº®¶~´Ï~§~£Ïqfj\!¹~§~£Ïqfh]¹!yWy' (L (L (L ʥ s£¶Ï%(L[~§~Ï£ÏqgjÏ^~§~Ï£ÏqfgÏ ~§~Ï£Ïqfj~§~Ï£ÏfgjÏ WÏ} [Ï} ʥ +ʥ +ʥ (L V}W} lÏ®Ï~§~Ï£ §®~`cfgÏ[Ï2.L8ÅÏ *2L
;2, _¡©¶§F Ï# k§©«Ï ¡ £«Ï « ~«Ï Ï ´§½Ï «Ï¡Ï « Ï A #«Ï ¡ ±§Ï ~ ¥¥~§©Ï ~ «Ï~©«Ï ° Ï « ¡Ï « § ~¢¡£«Ï Ï£±§Ï ©«¡«Ï «©Ï ¡Ï«·£Ï ¯Ï¡©«¡«Ï ±§'«©Ï ÏĐʥ£«§Ï ¶£§©Ï «Ï¡±§Ï ~¡Ï£¡½Ï Ï£§Ï ½Ï±©¡Ï £ ¡Ï «Ï £ §Ï « §Ï ©««Ï «©Ï §©¥«´½, Ï z§£§Ï ¶ Ï £¡©§Ï « §Ï ~©©, Ï d~©Ï«Ï #zÏA$«Ï¡±§Ï©Ï£§Ï½Ï£¡½Ï£¡Ï«'Ï z¡Ï ¡ ±§Ï Ä £ Ï©±Ï A $«Ï ¡ ±§©Ï©Ï~§½Ï E ' Ï d~©Ï6«Ï FÏzÏ A$«Ï ¡±§Ï ©Ï£§Ï ½Ï«¶£Ï©«¡«Ï«©'´¡Ï Ï «©ÏÏ~¡Ï k§©«Ï ¡ ±§Ï £ Ï©±Ï A $«Ï ¡ ±§©Ï£§Ï½Ï«¶£Ï ¶§Ï#ŐőʥÏVSTÏEÏ©Ï ÌM L ͪ¤Ï rº«Ï « Ï ¡ ±§Ï £ Ï©±Ï A $«Ï ¡ ±§©Ï £§Ï ½Ï 2 Ï~¡Ï ~ Ï ´¡Ï «Ï ¶ §Ï # ST ÏEÏ©Ï =>o
z§£§Ï«Ï«£«~Ï¡±§Ï£Ï©±ÏA$«Ï¡±§©Ï£§Ï½Ï«¶£Ï©«¡«Ï«©Ï ι?2E¹Ï6?ÏGM926?' d~©Ï«Ï¡±§Ï DʥzÏA$«Ï¡±§Ï©Ï£§Ï½Ï«§Ï©«¡«Ï«©1Ï k§©«Ï £ Ï©±Ï A %«Ï ¡ ±§©Ï£§Ï ½Ï«§Ï ´¡Ï «©Ï Ï~¡Ï k ¶§Ï ŐőʥÏVÏVÏk0EÏ©Ï QU
G ),H. "<$«Ï M ¡±§©Ï£§Ï½Ï2Ï~¡Ï«·£Ï´¡Ï«©ÏÏ~¡ rº«Ï « Ï ¡ ³§Ï £ Ï©±Ï A ϶§Ï STÏÏVSTÏEÏ©Ï A¤ ' G (+H. z§£§Ï«Ï«£«~Ï¡±§Ï£Ï©±ÏA&«Ï¡±§©Ï£§Ï½Ï«§Ï©«¡«Ï«©Ï #
QU
#; ¹ÏE2Ï,ʥÏÎÀÏA2ÏGMEB62 l¡Ï«Ï~¡©¶§Ï©ÏEÏ,ʥ626?ÏÏEB62Ï # +O
#
#
Muʥ ŀʥ ʥ6ʥ-2ʥʥ &ʥ6ʥàʥ-( '!ʥ#ʥLoʥ Vʥ '!ʥ ʥ /ʥ / / àʥ-!( '! 'ʥ!(!/:ʥ ' ¬ !( {ʥ ʥ ʥ ʥ ʥ ʥ
9
'O
;
@2ʥʥ !(
)0; ʥ2ʥʥ!ʥ ʥʥ!ʥ 2ʥ lo
9M
/ʥ. , / h #ʥ 'ʥ + àǵʥ U !ʥ0ʥ "2ʥʥ6àʥ-( ' "ʥà 'ʥ@ 6œʥ 2ʥ œ ʥ V ;2ʥʥ!ʥ0ʥ '2ʥʥ6ʥ 2ʥ ʥʥ #ʥ U !ʥ0ʥ +2ʥ 6ʥ 2ʥ ʥ #ʥ U !ʥ0ʥ -2ʥ 6ʥ 2ʥ ʥ +#ʥ Uʥ2ʥʥ ^ʥ& ʥʥʥʥ 2ʥʥʥ ʥ ʥ!ʥ0ʥ +ʥ 6ʥ0ʥ
#ʥ #
HDʥ " mʥ Æ!Ɗʥ!Ƌʥ!ƌ ƀʥ:ʥ
F%IAƁÆ!ƍʥ!ĩƎ#
ʥ ʥ ʥʥ#:M ʥʥ#:@ 2ʥʥɔʥeʥʥ!/Ȅʥʥʥ^$ ʥʥ · ! ȓʥ ʥʥ!/ʥʥ!±Ţʥʥ#ʥmʥ ʥ ʥʥ#:ʥ&2ʥʥʥeʥʥ!/ ȅʥʥ ^$ ʥʥ!ǷȔʥ ʥʥ!/ʥʥ! ±ʥŢʥʥ ʥǥ#ʥ@ʥʥ;ʥʥʥɇʥʥ Ɏʥʥ!Ƃʥʥʥ^$ ʥʥ ʥ!Əʥʥ!ĩĪʥ ɉʥ#:0ʥ#ʥ Lo
&MID -M
Áʥ%:0ʥ2ʥʥ &ʥ ƃÆʥ!īʥ!Ĭ Īʥ:
%IB
%"CƄÆ!īʥ!ĬƐ
" /ʥ ±ʥ ǔ 0ʥ ʥ7M [ʥ #ʥ [ʥ[ʥ
" 0ʥ 5!/ ɹɼ 4 !/ ə ƥ!/ ȥ ʥ Y
{E
o
ZE
o
ʥ
Lo
'*MEMʥ ʥZ`ʥʥ Iʥʥ:#@M ʥ:ʥ"Cʥ #
?
ʥ` Iʥ ʥʥ`e`ʥʥ!±ʥʥʥ^$ ʥʥ r /űʥ ʥʥ!/űʥ dDS ho
M
Iȇʥ ʥ AIʥ `ʥ5 ʥCʥʥ#rʥʥ`ʥxfʥʥ`.f.fƦȦʥʥ &ʥxfʥÏʥ.fʥ ʥxfʥÏʥ.fIJěʥ GʥxfʥŘʥ.fƧě(.fƑʥ ʥʥxfʥŘʥ5ʥ ʥ/ʥ ʥ(5 /ʥ ʥ`ʥ&ʥxfʥǢC5ʥ ʥ ʥ# Z`ʥxfʥņC5Ǯʥʥ (5C5ʥ ʥ ʥʥ cʥxfʥņC¬5#ʥLo ʥ xfʥǨCCĶ5ʥʥC5ʥ,ʥ ʥ ʥʥxfʥǣ ʥF
+O
X
*O
r
M
IŌʥ
AcIʥʥIc./QQÌʥ ʥ./QQĊʘʥVG$$ʥ] ʥ./QQÌʥçʥ!¯!/! °!±! ·!Ŋ!ǽ!ÌʥI ]ʥ!Ůƒʥ Iʥc
`ʥʥʥ ʥCʥʥʥʥ"ʥ'ʥ+ʥrʥ iʥ Íʥ.ɢōʥk:ʥ Cʥr#rʥǦC V`ʥ'Eʥ ʥM C+ʥʥG`ʥ ʥ!¯ʥ ʥ ʥ ʥIʥ IʥIʥ ʥC+ʥ GIʥʥ ʥG`ʥ ʥ]ʥeIʥʥʥ rʥ HʥCʥ "Eʥ@ C+ʥ@ ʥ "Eʥʥ &ʥ!/ʥçʥʥ ʥIʥ Iʥʥ ʥC+ʥGIʥ ʥG`ʥ ʥʥeIʥʥʥ ʥ ʥʥ`ʥʥʥCʥIʥ ʥGʥIʥ IʥIʥ ʥC+ʥGIʥcʥ ʥG`ʥ ʥʥ´Iʥʥʥ ʥ ʥʥ`ʥʥʥCʥʥIʥ r V I;ʥc`ʥCʥ "Eʥ,ʥʥ Eʥ@ C+ʥ@ Cʥ "Eʥʥʥ Eʥʥʥ ʥʥI ʥ!Ǵʥʥ'rʥa;ʥI$ ʥʥ IGʥ ʥGʥʥ
'"ââââʥʥʥʥeIʥ $$ Iʥ ʥʥeɑʥʥʥ./QQ·ʥʥʥGIʥ I II ʥʥ`I c
ʥIIʥF
{
{
+O
{
{
{
{
{
{
Cʥ
{
{
{
{
{
{
ʥ#
HDʥ ʥ Áʥ 5::ĆQʥx ʥĆQʥ ʥ: ¡t Vʥ.ʥ::ĆQʥx6 ʥ & W % |ʥ 5: 0 . ,ʥdo ʥʥ ʥdo ; #ʥ Vĝę 2ʥʥbʥÍʥ: ¬ 5:5ʥ & ʥ( 5:0ʥbʥ,ʥ6o ʥʥ ʥ;
ʥ
Dey\O~X =O`XtOeSOp D~SeX Efz]PY>PbYuPfTPqAqufPV E>A E;;Q !"' $! ' !% ' ' '
$%"&' ' % ' ' ' ' ' !# # ,o !# # # # # " # ! # # # # o
# # # !
I\} Nx Wq[ Ox X\uscq}|sq |c[\| BCx Wq[ BEx s]}xcWqak\ BCEx x\|u\Y}c\k F]
HNx EOx
?x % NBx sOBx x |bs
}bW} NxOx uW||\|}bxsab}b\Y\q}xsc[s] BCEx Bcq[WkkuWcx|s]us|c}c\cq}\a\x|rg }bW}|W}c|` }b\\wW~csq ,xt)fx x%xh r! I \} Bx X\ Wq r\k\o\q} |X|\} s] n%x)x xx)##5px
c}b }b\ uxsu\x} }bW} }b\[c] ]\x\qY\ X\}
\\q Wq }
s qoX\x| cq Bx c| qs} W uxco\ qoX\x Bcq[ }b\ kWxa\|} us||cXk\ Wk\ s] r Bcq[ W |\}
c}b }bc| qoX\x s] \k\o\q}| Ms}\, %x c| qs} W uxco\ qoX\x )
I\} TxUx# L #x |Yb }bW} Tx xUx x # x\WkqoX\x||Yb }bW} l&l*x xl0
/O
/O
%x Qbs
}bW} c]c] l&xl* x xxl0x Wx\ us|c}c\ %x }b\q
TlXx xUl&x WxTlw xUl*x WuutxTlq xUl0x Wx JK %x
-
Eq WqWxYb\x Ysou\}c}csq }b\x\Wx\ ,#x Ysq}\|}Wq}| Tb\}Wxa\}c|[cc[\[ cq }
s sq\| 9 bc} W} sq\ %x c| W
Wx[\[ %#x uscq}|
bck\ W bc} W} sq\ )x c| W
Wx[\[ 1 uscq}| Msuscq}c|W
Wx[\[ ]sx Woc|| AWYbYsq}\|}Wq} |bss}| %2x Wxxs
| 9} }b\ \q[s]}b\ Ysou\}c}csq |}W}c|}cY| |bs
}bW} osx\bWq 1#x s]}b\Wxxs
|bc}sq\ ) Tb\qoX\xs]Wxxs
|}bW}bc}sq\ %x Wq[oc||}b\}Wxa\}Wx\\wWk Oxs\}bW} }b\x\Wx\}
sYsq}\|}Wq}|
c}b}b\|Wo\|Ysx\
((o
Cey\O~X =O`XtOeSOp C~SeX Efz]PY>PaYuPfTPqAqufPV E>A E;;Q !"'$! ' !% ' ' #!%$! #'
I\} Gx X\}b\od[usdq}s] BDx QdqY\ M A %xOx kd\|dq }b\|\ao\q} DGx I\} Ix X\ }b\dq}\x|\Y}dsq s] CGx Wq[ NxOx ; L\q\lW| Tb\sx\oWuuld\[ }s}b\kdq\ NxOx Wq[ }xdWqak\BCxGx
GIxx CNxx BOx @x% ICx NB OGx
Tb|
CIx CNxx BOx DO x BO %x IG NBx OG OBx OG OxB DOx )DG DOx DOx OGx OGx )OGx ?x ?x)r OGx Tb\x\]sx\ Ix d|}b\Y\q}xsd[s]BCDx 1O
N
O
V\bW\ g*
2O
^
%x os[ ,x Tb| gx?x,]x x%x sx ,]x x)x ]sx|so\qsqq\aW}d\dq}\a\x
]x d gx?x,]xx%x 9]\x|doukd`dqa
\bW\)fx?x,] *x x)]x?x],]x x)x Tb|]x Wq[ ,]xx)x Wx\Xs}bus
\x|s] )x F} d|Yk\Wx}bW} ]x?x )x d| W |sk}dsq Wq[ ]x @x %x d| qs} F] ]x?x)Qx
b\x\v345 )x }b\q ,]x x)x?x ),x )Q`x x%x d|qs}Wus
\xs] )x W| ,xrx)Q axx %x d| s[[ V\bW\sq\ |sk}dsq- q h ( g ?x % dd gx?x ,]xx)6x 9aWdq
\bW\ )f ?x,]*x x/]xx%x?x,]xx%]xx%x Wq[Xs}b ]xx% Wq[,]x x%x o|}X\us
\x|s])x ;s}b]x?x %x Wx\|sk}dsq| Vb\q ]x?x g @x
bdYb d| qs} W[od||dXk\ Bsx ]x L %x
\ bW\ ,]xx%x @x )]x x]xx%x L )]x x)x Wq[ }b\x\]sx\ %]xx%x L ,]xx %x L )x]xx%x D\qY\ d] ]xx%x ?x )Qx ]sx |so\ us|d}d\ dq}\a\xv }b\q %&
)R *x L ,^x x%x L )R &x Wq[
\ZsqZk[\}bW} ,^x x1 ZWqqs}X\Wus
\xs])x Tb|}b\x\ c|sq\ |sk}csq cq }bc|ZW|\. r p 1 ,x /O
\} :X\Wq r\k\o\q}|X|\}s]o %x)x xx)$$5px
c}b}b\uxsu\x}}bW}}b\[c^\x\qZ\ X\}
\\qWq }
sqoX\x|cq: c|qs}Wuxeo\ qoX\x Bcq[}b\kWxa\|}us||cXk\Wk\ s]r Bcq[W|\}
c}b}bc|qoX\xs]\k\o\q}| Ns}\, %x c|qs}Wuxco\qoX\x x H] r Io : }b\q r x # |: )x,x1x( 9osqa r x%xr x/xr x2x W} os|} sq\ ZWq X\ cq : Tb| Wosqa Wq 4x Zsq|\Z}c\ cq}\a\x| W} os|} )x ZWq X\ cq S D\qZ\ G : G 01 ) b)$$5"4cx i 1$/x QZbW|\}c| o/^x x% 7 x^x i $x%x! ! ! xx1$)p x /O
)
V\ac\Wuxss]s]}b\a\q\xWk ZW|\ >sq|c[\x }b\\uWq|csqs]
TlXx xUl&x xWTlwx xUl*x W r r r x Tl< xUlhx Wx Y
Tb\}\xocq T exW _x
b\x\# xh x^x
$O
rc|
Y
Tb\x\Wx\Wk}sa\}b\x , h[ x }\xo|cq}b\ ooW}csq V\Zbss|\ # ]WZ}sx|]xso
bcZb
\ }Wi\ Tl= x Bxso }b\ x\oWcqcqa r # ]WZ}sx|
\ Zbss|\ h }s }Wi\ }b\ }\xo| Ul-x < |oo\}x }b\ qoX\x s] }\xo| Zsq}Wcqcqa l>x c| W Zsq|}Wq} W| c| }b\ qoX\x s] }\xo|Zsq}Wcqcqa }b\}\xoSY vx Tb|
b\q}b\}\xo|cq }b\|ooW}csq Wx\ok}cukc\[ }sa\}b\x
\a\} l&l*x! S x h Rx 1 ]sx |so\ # Bsxsx uxus|\ c}c|qs}q\Z\||Wx }s Y Y Zsou}\ # Hq ]WZ} # i ) ,1 x h[ x x JAZ xwh h , F[ x < }b\ 9LCL cq\wWkc}
\bW\ $O
D\qZ\
Tb\ qoX\x s] Wxxs
| }bW}}bW} bc}sq\ %x c| A ,$x %2"/x %)$x H]Zsq}\|}Wq} bc}|sq\ %xTYx }co\| sq\)xUYx }co\|Wq[oc||}b\}Wxa\}WYx }co\| }b\q}b\}s}Wk|Zsx\c| %$TY 1UYx 1TY 1TY UY x 1TY 1%2 WYx 4$ 1TY WY rx Quus|\}b\|Zsx\|Wx\Wkk [c|}cqZ} }b\q }b\ ,$x qoX\x|TY WYx o|}WkkX\[c|}cqZ} < }b\uca\sqbsk\uxcqZcuk\ bWk] s] }b\|\ ,$x qoX\x| Wx\ \c}b\x us|c}c\ sx q\aW}c\ V\ Zsq|c[\x }b\ us|c}c\ ZW|\ Vc}bs} ks|| s] a\q\xWkc} k\} TYx WYx L $x ]sx # %x ! xx%1x Tb\q TYx WYx JK Tb\x\]sx\ TYx # # D\qZ\ T&x xrxrxrx xT&0x HI %)$x <}T&x xrxrxrx xT.jx A %)$x Wq[
\bW\W Zsq}xW[cZ}csq Tb\ q\aW}c\ ZW|\c||cockWx -
LO
/O
/O
$O
/O
/2x
/O
>CBHCLB@PR'R;NHCR #5R
5'#R
:HLMJNAMEIHLR MIRAIHMCLM@HMLR
"'-&22 ,R %+'*#"'2
&*2/#+&2 "'-&'2 "2*2 "'-&2 '*2$,2"2'2*2$$$&*2+ '2 #/#+&2"'-&'
%
#2 '*$'2&2 "2 *#2+'*2/#+&2"'-&'2
*
2%+'*#"2&&'2 R !&2
,
#2 + *#&'2&2 #-2
?A ?A ?A ?A +'A ';34'55,20A A5,0A %25A %25A %25A %!0A#'A 5,/3.,)'&A!5A25,0A 2 :+'4'A 2 !1&A $!4'A325,7,9'A,07'*'45A:,7+AA 2 A ,0&A7+'A9!.8'A2(A A2
'7 A AAA6A!0&AA#'A 5,;A32,075A20A!A%,4%.'A,0A7+,5A24&'4A58%+A7+!7A77"RA 7"7&R
7&7+RA 7+7/R !0&A7/72R 727R :+'4'A 77"R &'027'5A7+'A !4%A.'0*7+A 2(A7+'A !4%A 77"R '7%A 7A ,5 !.52A -02:0A 7+!7A <77&7/RA @A ,0&A7+'A5,>'A 2(A =7+727"R ,0A&'*4''5
$
AA A #'A A&,57,0%7A 32,075A 21A 7+'A 5'*/'07A 89R 2(A!A74,!0*.'A 689R :+'4'A 68RA 69 A A 9!.8!7'A 7+'A 58/A R R
)
'7'4/,0'A 7+'A .!4*'57A 9!.8'A 2(A ?R (24A:+,%+A
'7A
-
'7A (
R R O FFOR2 2!O 21O .1
O
R
AO A
2 (2 2 2 .2 . 2 2 0 2 2) #'A !A32.=02/,!.A,0A OR:+'4'A 2 12 2 2 2 !4'A
%2057!075A!0&A"6A A
+'0A &,9,&'&A#=AO A O A O A O A !0&A O A
(<A.'!9'5A !A 4'/!,0&'4A 2(AA A A A !0&A A 4'53'%7,9'.=A ,0&A7+'A 9!.8'A2(A(A
0 R
,0&A7+'A 9!.8'A 2(A
KDG4QR KDGRQR
RKDGR4QR
(3R
'7P7NE@G7]P>7]GQE17N]J8] 5@=@P]KJO@P@R7]@GP7=7NO] OQ4>]P>/P]/;7N]57B7P@G=] /GV]JG7]5@=@P]P>7 N7E/@G@G=]]5@=@P]GQE17N]@O] 6@R@O@1B7]1V]
$
*P] @O] =@R7G] P>/P]
S>7N7] 0] /G6] $ /N7] N7/B] GQE17NO ] )@G5] P>7] E/T@EQE] KJOO@1B7
R/BQ7]J8]0 3
+7P] $%&] 17] /] PN@/G=B7] S@P>] O@57O] $%]
]%&]
] /G5] $&
] $] QG@MQ7] 4@N4B7] 4/G] 17] 5N/SG]
PJQ4>@G=] P>7] O@57] $&] /G5] P>7] B@G7O] %$] KNJ5Q475] /G6] %&] KNJ5Q475] +7P] '] 17] P>7] 47GPN7] J8] P>@O] 4@N4B7] )@G5] P>7] R/BQ7] J8] %' ]
#$
*8] #
$
+7P-#] Z]]]
]][] '7P7NE@G7] P>7] GQE27N] J8] R74PJNO] U]W]
P>7]R/BQ7]J8]
S@P>] U]W] -]
OQ4>]P>/P U]"] /G5W"] "]
+7P] 8H] 17] P>7] GQE17N] J8] ,O] @G] P>7] 574@E/C] N7KN7O7GP/P@JG] J8] P>7] KJO@P@R7] @GP7=7N] H] )JN 7T/EKB7]8]
]/G5]8]
)@G5]P>7]R/BQ7]J8
8] 8] 8] 8]
OQ4>] P>/P] U] A] R/BQ7O] J8] !
$
*P] @O] =@R7G] P>/P] $ @O] /] KJO@P@R7] @GP7=7N] GJP] 7T4775@G=] ].>7N7] /N7] GJ] G/PQN/B]GQE17NO] U] /G5W
+7P]-]
] )@G6] P>7] 6@887N7G47] 17PS77G] P>7] E/T@EQE] /G5] E@G@EQE] KJOO@1B7
Z]]]]$ ]][] *G]7/4>] J8]P>7] 8JCBJS@G=] OQ1O7PO] J8]-] Z[] Z]][] Z]]]]][] Z]] ]]]]]][]
P>7] OQE] J8] /BB] P>7] 7C7E7GPO] @O] /] EQBP@KB7] J8] ] )@G5] P>7] PJP/B] GQE17N] J8] GJG7EKPV] OQ1O7PO] $ J8]-] OQ4?]P?/P]P>7] OQE] J8] /BB] 7B7E7GPO] @G] $ @O]/] EQBP@KB7] J8] ]
$
$]
$ O/P@O:7O] P?7]N7D/P@JG] 8U8X]
8UW] ] 8U] W 8U] U]S>7N7]
U]W] )@G5]P>7] R/BQ7] J8 8]
+7P] Z0I[] 17] /] O7MQ7G47] J8] LJO@P@R7] @GP7=7NO] OQ4>]P?/P]0]
$ '7P7NEF7] P?7] R/BQ7] J8]
] 0]
]/G5]8JN] H] ]
'7P7NE@G7] P>7]GQE17N]J8]S/VO] J8]P@B@G=]/] T]N74P/G=B7] 1V]P@B7O] J8] O@Y7]]T
)@G5] P>7] GQE17N] J8] 5@=@P] KJO@P@R7] @GP7=7NO] OQ4>] P>/P] P>7] 6@=@PO] 8NJE] B78P] PJ] N@=>P] /N7] GJG\ @G4N7/O@G=] (T/EKB7O] J8] 5@=@P] GJG@G4N7/O@G=] GQE17NO] /N7] ] /G6] !] $G 7T/ELB7]J8]/] GQE17N]P>/P]@O] $ GJG@G4N7/O@G=] @O]]
$
*&
!4 *4 &(*4 $&4 "- '4 ()4 * !4
, #4 *4 #&
4
$#(*04!*&(4 !4 $, (4 ' +'4 , "4 , ,
*&
!4 *4 (*4 #"), ",4 #,)-4* *4 )*+''',
1&4 !4
,
#&4 !34$#(*04 & 4 ". &(4 ,
,
*#4, !/(04 *44*4 $ *34* *4 *4 (*4 *&4 #4 *4 !/ &(4 &4 #!(/*04 !4*40 /4 #4 #*4 !#*(4 *4& *(*4!*&4 ((4* !4#&4%/ 4*#4 24 !/
&(4 &4 & !#34 (*4
0 / *4
!,
*&
, ", * !4
,
,
!4 *4 &(*4$&
4*4!*&(4
,
4 *#&4#4 *4 (/
4
! ",
,
*4 &$, ", 4 4(%/!4#4 & 4 !/
&(4(/4* *4
$, #&4 0&34$#(*04!*&(4 *&
, ,
, $", $, , ,
!4 *4 0 /4 #4 4 $ , $(%,
,
, !
&4
$ " " "" " " " " $ " $ " " " " " " " " " " " " !" $$ $ " " "
!" " " " " " " " " " " "
" " " " "
" " "
" "$ " " " " " "
" " " "
" " "
"
" " " " " " " " " " "
$
$
$
$
$
$
$
$
$ $
$
$
$
$
$
$
$
!% ' #
## ## ## ##
! ! ! ! !#
# #
# !# # !# # !# !# # !# !# !# # ! !# # ! !# # '
! ! ! !# #
#
!# !%' ""'# # '
#
#
!% ' # !"' ! $' ""' ' ' '
" "#
"# ' '
" '
"# !#"' ' '
" #
" &' #!'
3
38<*7@
43897:(9@&@(07(1*@ (*397*)@&9@)3 <09.@7&)0:8@)*3(3)+3 08@,.) !3 <.0(.@08@&184@*6:&1@94@ *,/,.+3
".:8@
!3
".*@54<*7@4+@,.3 <09/@7*85*(9@94@
3
,.) *,.3 2,.+(3 !
3
".*7*+47*@9.*
8:2@*6:&18@$3 03 @(3 3
38<*7@ !03(*@ =@ @= @=@ =@@ @ =@ @@= @ 9.*@-0;*3@*6:[email protected])8@0+@&3)@431>@0+
!03(*@
@ 0+@&3)@431>@0+@'3 ".*@&'4;*@03*6:&109>@7*):(*8@94 = = ==@! %&3
!03(*@ ?=@= = @= =@@! %&3
!3%@$
<*@(43(1:)*@9.&9@9.*@1&7-*89@;&1:*@4,@=08@ @
3
38<*7 @ #*@.&;*
1 -1 1"1#3
33333
111"1#3
3 11 1"1 #3 3 3 33 1#3
3 11-1 33
3 3 3 11 -1 1"3 3
3
3
!4@9.&9@+@ @
3
3
$
'.4- 6 $
/"'6
/"'6
$
'.4- 6 06$
$#$ 6 .1 66'1&-6 56$0#'66 '6 6 4606 $#$ '6 $#$
-.)0#3$56 (0 6(6 0 &6-6 #3#.#%656 $ '60 #-6 #-'6 $ 6 #.6 %.(6 #3#.#$656$ 0 -(-6 6 6#.6#3#.#%656 $ 560 6.#$-6 -1&'06 6 $
! "$ !$-6 #3#.#%656 $ '6(0 -64(-6 %%6 0 6 ##0.6(6$-6('-1'06&(1$(6 $
$('0#'.6##0.60 06-6-0-60 '6 $ 0 '6('6'6.10-06$ -(&6 6 ##060(6 066'46 '1&-6$ '6$.0#..60 6 -,1#-&'0.6#'60 6,1.0#('6#6'6('%56#6$ (.6 #'6$$6 0 6 ##0.6-6 ('-2'06&(1$(6 $ #06 -&#'.60(6 ('.#-6 '1&-.6(60 6(-&6 ********6 4 -6+6
*******6
$ $ $ $ $ $ $ 56$0#'66##06 #'60 #.6'1&-6 46 060 6 '1&-6
*6 4 # 6#.6 #3#.#%6 56 $ #6'6('$56#+6 #.60 6 ##06$ (-6 $ (43-6
0 6 -.0604(6 ##0.6(6$&1.06 6 $.#'6 0 6'1&-6$!.6 $ ##0.6'60 06 '56'1&-6 46 06 56 $0#'66##06 #'6$ .6 $##0. 6 '6 0 6 (0 -6 '6 0 6-&#'#'6 $##0.6'66 $ $ (#.6(6.1 6'1&-.
#')''0%56$ (-6 $ ('.,1'0$56 0 -6 -6
$
'.3-7 7
'.3-7
070#7 $-&73$0#7'0-7 * '7-$1.7 0*1#7 0#7 0("'07 %$'.7 **+-*17'7 +-*17 07 0#7 +*$'0.7 *
*(7 !*-.+0$2&67 #(7 * * * * '7 0-$'"%. *(7 !*-7*'"-1'07 '7 #'7 "**"** "* 7#2 */7
*
*
7 '7#'7 / )
*
* *
*7 !(7 37 #27
* ** 3#-7 * '*0.7 0#7 -7 * 70-$'"%7 * 07 '7
&(* $( %(* )) $) %) &* 3#-7 * *
*%2$("7 37 "07 #* *(.$-$'"7 0-$'"%7
* #27
*37 #27 *** #1.7 37
7
'.3-7 7
$'57
$.77 -**07 * 70#7,10$*'7 ' 40
3#$#7 $.7 (*07 +*..$%7 #1.7
37 #27
7 .7 0#0
7 #1.
*
*-7
7
7
49>(8H H !+(8(H #8(H;>6H %#9(9H;6H %649,&(8H #9(H H 6: DAH BFH#4&H#9(HH 6: EH BF H 68H#9(H H
;+(8(H #8(H : H H
>#@9H#4&H)68H#9(HH ;+(8(H#8(H
>#@9 H (4%(H ;6;#0H 4<3$(8H6)H>#@9
49>(8H H (;:H D H H H HH FH #4&H:-H H D5H
68H H .H ,)H5H
!: )5HH.FH)68.GH H !+<9H
!': #4&H5H +#9H(?#%;1@H/H&,*,;9H,4H;+(H&(%,3#1H8(78(9(4;#;,64H ;+(4H (?#%;0@H
>(H +#=(H/.H&,*,;9H #8(H 464C(86 H !+<9H
"(/ )(0:
)(:
!+(4H ,;H,9H %0(#8H;+#;H
49>(8H 6;(H ;+#;H H :HH #4&H : )68H #00H 4#;<8#0H 4<3$(89H
# :
!+<9H
:
H <7769(H /HH H 36&H H6;(H ;+#;H
6:*: 36& 36&H $<;H
(4%(H ;+(8(H #8(H 46H 4#;<8#0H 4<3$(89H 6: #4&H 9<%+H ;+#;H !+(8()68(H 3#?D/F 3,4D/FH
#: H H36'HH
H
*:
36& H
H ,)H /H,9H #H 3<0;,70(H 6)H H
H
49>(8H H (;H
96!:
"H .H36&HFH )68.H H H 6;(H ;+#;H !4+:
H !,+:H H#4&H ! :H H (;H $(
;+(H 9(;H 6)H#00H 9<$9(;9H: 6)H !:9<%+H;+#;H 7&$:8:,9H #H3<0;,70(H 6)HH 6;(H;+#;H)68H#4@H: !:
:
8: 83" 3": 15%
( :
!+<9H : ,)H#4'H 640@H ,)H 2H 2:!-.: :2! :36'H H !+<9H ,;H,9H %0(#8H;+#;H
(#!'** F1,?,
*
*
*
*
*
*
*
*
*
*
",;*,L L L L #3;*,LF,LF(;AL<;9IL;<;,:=AIL@B)@,A@L F,L1(D,L
;@F,?L !3D,;L-G-L -GJLLLL-GL L J -GLLGL@
-GJLLLL-GL L J -LLJ
#B)A?(*A3;0 LF,L1(D,L -GLLGL
-KLLJL -
*<;@A(;AL 9,AL3AL),L4L #<L-GL GLL4 #B)@A3ABA,L)(*5LAB(A3<;LF,L1(D, GLL4JLL4L HL LLL4LL&GLLJLL4' GLL4LLG GJLL7LL8LL 6GLLJ GLLJL
GJLLLL4LLGLLJLL4 G 4LLG
4 L
4 4LLL GLLJ 4 4 4LLLLGLLJL L -
;@F,?L
"& $"&
"& * "& $ "&)*
"& $
?<:L1,?,L F,L@,,LA2(A
"&*
3@L(;L(?3A1:,A3*L@,>B,;*,LF3A1L.?@ALA,?:L
"&*
L(;+L*<::<;L+3--,?,;*,L L $1C@L
"& %*
L
L (;+LA1(A
"&* "*
* *
* (;+L
* *
"
"* "* "
"*
L
*
* %,LA1,?,-
&HQW9P#] +9S];]49] S>9] HUF49P] M:] W2ZQ]M:] SAEAH=] 2] ] Y] I] P95S2H=E9 ] &EQM] E9S] 49] S>9] HUF49P] M:] W2ZQ] M:] SAEAH=] 2] ] Y] I] P95S2H=E9] WAS>] S>9] SMN] MP] 4MSSMF] SWM] QOU2P9Q] AH] S>9] E2QS] 5MEUG] FAQQAH=] 2H7] E9S] 49] S>9] HUF49P] M:] W2ZQ] M:] SAEAH=] 2] ] Y] I] P95S2H=E9] WAS>] S>9] SMN]2H7] 4MSSMF] QOU2P9Q] AH] S>9] E2QS] 5MEUFL] FAQQAH=]
-9S] UN] 2] QZQS9F] M:] P95UPP9H59] P9E2SAMHQ] AHVMEVAH=] ;] ] 2H7] 4Z]
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
/AS>] AHASA2E] 5MH7ASAMHQ<]
] 2H7] ]
-MEVAH=;]P95UPQAV9EZ] W9] M4S2AH]
]
&HQW9P#] " '25>] 7A=AS] 4AH2P[] Q9OU9H59] 5MHS2AHAH=] 9Y25SEZ] HAH9] ,Q] 2H7] Q9V9H] ]Q] 52H] 49] F2S5>97] UHAOU9EZ] SM] QU5>] 2] ] 7A=AS] AHS9=9P] MP] ] 2Q] :MEEMWQ#] '25>] ] WAEE] 49] P9NE2597] 4Z] 2] 7A=AS] :PMF] ] SM]"] AH]S>AQ]W2Z#] S>9] HUF49P]M:] ,Q] SM] S>9] PA=>S]M:]2] N2PSA5UE2P] ] AH7A52S9Q] S>9]V2EU9] M:] S>9] 8B=AS] (MP] 9Y2FNE9] ] .>UQ] P9OUAP97] HUF49P%]
!!] 2H7] ]
"""" ]
&HQW9P$] ]
AQ] 2] NPAF9] S>9H] ] -UNNMQ9] I]
] W>9P9]925>] AQ] 2] 5MFNMQAS9] S>9H] -AH59]
:25SMP] MH] T?9] PB=>T] AQ] =P92S9P] S>2H] 5MHSP27A5SAH=] S>9]:25S] S>2S]3K] ] AQ] 2] NPAF9 ] .>9P9:MP9] I] ] FUQS] 49]2] NPBF9] .>9] E2P=9QS] NPAF9] I] QU5>] S>2S] ] ]AQ]] )MW9V9P] ]%]]] ] Y] !"] X?C6@]
JMS] 2] NPAF9 ] -BH59] ]
OU9QSBMH] BR]
] AQ] 2] NPAF9] HUF49P] S>9]2HQW9P] SM] S>AQ]
$)-( 0 ,
!!
, &
,
!
!
,
%*0 **0 &,, %,
,
!
! !
!! !
,
*, 0 .*0 )*0
! !
,
,
!
!!
/0 *0)#0 ()%$$0 &, !,
, 0 !#0 **0 &,! , +&&%)0**0 &, ! , *$0-0 #+)*0 ,0
, ,&,, &, ! ,, , , , -0)0 )+(0 $0 %+(0 "#0 **0 &,! , )0*(+0 ) 0 )!!0 $.*0 )%-0**0 #,)0 *0!)*0 &%))!0 %+$0%(0 &, %$)(0 *0 $+#()0 +( , ' , !
!
-(0 " , , , ,, , , (*0 %#&+** $0/ !")0 $,
!
!
, ! , $0 ', )0 %)$0 ! ! !
*%0 0(*((!/0 "%)0 *%0 , *)0.&())%$0 *)0 (*((!/0 !%)0*%0 ,
$)-( 0 , 0 $+#(0 %0 )!*%$)0 )+0 **0 $%0 *-%0 %0 *0 $+#()0 (0 %$)+*,0 , 0 $+#(0 %0 $(/0 )'+$)0 %$*$$0 , !,)0 $0 , )0
0 $+#(0 %0 )!*%$)0 )+0 **0 .*!/0 *-%0 %0 *0 $+#()0 (0 %$)+*,0 % $(/0 )'+$)0 %$*$$0 0!! , , !,) $0 ,)0
0 $+#(0
0 $+#(0 %0 )!*%$)0 -*0 .*"/0 *-%0 )*)0 %0 *-%0 %$)+*,0 $+#()0 +*0 $%0 *(0
0 $+#(0 %0 $(/0 )'+$)0 %$*$$0 , ,)0 , ,)0 $0 ,
$+#()0 (0 %$)+*,0
)0
0 $+#(0 %0 )"*%$)0 -*0 .*"/0 *(0 )*)0 %0 *-%0 %$)+*,0 $+#()0 +*0 $%0 *(0 $+#()0 (0 %$)+*,0
0 $+#(0 %0 $(/0 )'+$)0 %$*$$0 , ,)0 $0 ,)0 ,
,
#%
'
#%
% %% %
% "%%
'
'
'
%
'
!'
' '
$% #"'
%% % ' ' % % % ! % #% '
% %&' ' #'$' % % % %
'
'
(
$% ( % % #% % $% #%
% !
#(
"( # %(( ( % % #% % ( ( " (
$% $% % % !% "% % % ! #( #'$(
(
"(
%(( ( ( " (
( ((
(( (
$% (
% "% % % "% "%
( ( ( ( ( ( ( & ( ( ( ( (
% (
% "% % "%
( % ( ( ( ( (( ( (( ( ( (
(( (( ( (
% (
% % "% % %%%"% (( (
% %( (% (
(
(
"( $$ ( $&(
!#( % % ( '! ( ( # # !# # !"# !"# #
#
# # #
"&
& ! "& & & & & #!& " & ! & &
& ! & & " #( ( #( #( ( #( ( !( ( ( $!( % #!( ( ( (
( !( ( !!(
%%% ( % !#%'( !( ##( (
( (
##( #( #( !( ( &( ##( ( ( ( ( ( $( !( ( $( &( !( $( &( #( ( ( #!( #! !( %!( %( ##( #( ##( !!(
(#!( #'
'( ( $!( ( ( !#%( #( ( % (
%
%(( '(
% % % % % % %( ##( % !( ( #( ( ( % $& ( % !( ( ( ( '( ( % ( % %
( #( !#( !##( % !$( ##
%
!% % %&
% % ( ( !#%( ( $!( ! % %!% !$( ##
!% $% !%% %
( ( #!( ! %"% ( #% &#( % & !% & "% & #% "$( ##
!"%
( # % !#%
( " % "#%
( !%
( %
' ! " ' "$' ' # # '% ' '
'#9 %&4/#9 1'-619 '*$9 -$9 3'#9 /'-+ 419 6'(!'9 (19 ,9 (1-1!#*#19 3/(,&*#9 6'#/#9 9 9
/#9 .-(,319 -$9 3,,!79 -$9 3'#9 !(/!*#9 3-9 3'#9 1("#19 9 9 ,"9 9 /#1.#!3(5#*79 -3#9 3'39 9 #9
9# 9
9# 9
,9 ./3(!4*/9 # # ! #
'419 9 !## # 9 #,!#9 3'#9 3/(,&*#19 99 ,"9 99 /#9 1(+(*/9 39 $-**-619 3'39 9 99 # 9
(+(*0*79 -,9 3'#9 *-6#/9 '*$9 -$9 3'#9 /'-+ 419 9
9 9 '#,9 9 # 9
,"9 '#,!#9 3'#9 3/(,&*#19 9 ,"9 9 /#9 1(+(*/9 '419 9# 9 (,!#9 9 )19 ./**#*9 3-9 9 (39 $-**-619 3'399 (19 ./**#*9 3-9 9
#39 ### # #**9 1'-69 3'39 3'#/#9 /#9 (,%,(3#*79 +,79 # 14!'9 3'39 # !-,1(131
-$9 -,*79 3'#9 "(&(39 # (#9 # ### ## #
')19 7)#*"29 #
#
+-"9 # '419 ,79 .-1(3(5#9 (,3#/9 -$9 3'#9 $-/+9
+-"9 # (19 (,9 3'#9 ),!#9 #
# # " # #
# ,"9 #
+-"9 # $-/9 ,79 '419 3'#/#9 /#9 (,%,(3#*79 +,79 .*),"/-+(!9 ,4+8
#029
#
6'#/#
+-"9 # 6#9 '5#
#9 '5# # #### # # ### #
) ! ) %) !)
=F"F =F$F -> #F ==FF"->F*F$=F"F %9 # ==F"F ->="=->#F*F$=%9 #F =FF"=F$F -> #F="=->#F+F$=FF"%9 #F F$=%9 # ) ! ) ") %) ! )
$1= ->F4 ==F"F F
1=F+F"BF$F BFF F=7F +F$:F F"F
%)
>F >F $1=F*F0=6B6BF+F 0="6B6BF
)
%) ) ! !)
=F ! ) "
==FF"F $1=F " ) ->F "F ) $F ) ) ->F ) @55F
) ) !) ) )
1=F ) @55F
=F
) !) ) )
@55F ") ) @55F
= ) ) =$BF ) =F "
> %F 1=F 0F=6BF 6BDFBF ) =F
" )
->F
) $
=
&F ) ) =FF"!$BF ) " F #
> # %F 1=+F 0F 8,#F " )
->F
"F BF F 4?F <@5 =;F
) $
!) !) ! !
CAF2EF&2#2'"F*2/ F2E "'&2#2'F" +F2F# F 2F F'F"FF2F#F 2F 'F%F F 2F# F F2'F "F F&CF#F2'F2F#F 2F 'F "F *FC#F F2' "3C#C'%FFC#F "F C'F !) ) # ) !) () ') %) !)
C#FCF 'F+F"F
C#FFC'F+F"F ) C#F+F 2'F+F"F "
) %) # !) $)
) !') ) !$) !) %) "& ) !) %) ! !) !) %) ) )
(&F
$ $ $ $ $ $ $ $ $ %*.
*..* . $
%*..%*. %.
$ #$ $ $ *.+.,. $ $ $ $ $
"$ $
.
.
*.+.
*.,.
+.,.
.
$
. . *. .+.
.,.
$ $ $ $ $ !$ $ "$ $ $
*+..*,..+, . . $ *. +. , . $ *.+. $ ,. $ !$ $ !$ $
*+ *, +,!. ()$*+, . $ !$ $ *+..*,..+,. $ $
$ *.
&*+,. $ $ $&. .
$ #$ *+,. !$
. . . . " '. ,. +. *. *+,.
.+. .,. $ !$ $
$ $ $ *.
. $ $ #$
. . . *. +.
-
$ $ $ $ $ $ $
. .
. . +. ,.
. . . +.
+. $ !$ $ $ #$ $ $ $ ,. $ . $ . $ $ . *. .+. .,. $ $ $ *.+.,. . . $ . . $ . $ . $ $ #$ $ $ #$ $ $ *.+.,. . .
! $ $ $
.
YxrcmTcumcxgc}V}xch YTV5)); +%!&(-*!&%-
«ª¯¹ *ǎ¯¹
; Ô ǎª¹
$'&(*%*-
'A=J5021M )=AMG42M;I:G5>:2M/4=5/2M@I2DG5=>A=>A5-G2M.I..:2M.2:=KML=IAM-DM-A2M<22020MG=M8IDG7ML=IAM-
- -- - - - - - - ---
6¹
+"*!'"- &!- +)*!&%)>¹
UǎQ7ǎêǎ ǎǎ ǎ %ūǎ %ŪǎQ ǎ×ǎQ ǎ8} 'ǎ 3ǎ +ǎ .1ǎ 9ǎ
O¹
x8HQǎ 'ǎ 3ǎ +ǎ .1ǎ 9ǎ
[¹
M
=
;ǎ
^
ǎƖQ ǎǎHǎǎ® .
&;ǎ ǎ &;ǎ ;ǎ &ǎǎǎ ƉQǎǎǎHǎ
ǎ #ǎ *;ǎ ǎ ǎ
>ǎ8Q ǎQǎ*ǎQ11u¹ M;ǎQ21u¹ 8ǎ 8H8 ǎ7ǎ#"ǎ8 'ǎ 3ǎ +ǎ .1ǎ 9ǎ
f¹
ǎMLM
>ǎQǎǎ8QǎQǎǎQ ǎǎQ 8Qǎ8ǎ ǎ8ǎ .'¡ǎ .3ǎ +ǎ .1ǎ 9ǎ
b¹
%ǎ ;ǎ ǎ ǎ ǎ
ǎ "ǎ ǎ #ǎ ǎ
ƃğQǎǎ±ǎǎ Qǎ 8QǎQ ǎğǎ%ǎ Qǎ 8Qǎ8Qǎ8ǎ Sǎ UǎSǎ±ǎ
8Qǎǎǎ8ǎ SǎǎQ ŁǎQǎǎǎǎ8ǎ QBǎiǎ 8ǎǎ887ǎǎǎǎHǎǎǎǎ 8QǎSQǎǎ8ǎ*ǎ Q}ǎ # %ǎ 3ǎ ǎ ǎ +ǎ %ǎ ǎ .1ǎ & %ǎ ǎ 9ǎ ǎ 'ǎ
J¹
l¹
rǎ eǎ ǎ /¹ ǎ rǎ eeǎ ùǎ $ $$ ǎ$ ¬rǎ
ee¬ǎ eǎĐǎJǎejǎ eeƼǎ¬ǎ$ j}ǎ
¹
$
!ǎ
$
$
'ǎ 3ǎ +ǎ 1ǎ 9ǎ p¹
>rǎǎeǎrǎeǎǎrǎaǎ!ǎ+2 4 e 'ǎ 3ǎ +ǎ 1ǎ 9ǎ
t¹
#ǎ ¥ǎ aǎ a"ǎ a#ǎ
ĐǎJǎeǎeǎRǎIǎÏǎǎrǎǎrǎ<eǎRIÏǎ\ a¥}ǎ 'ǎ 3ǎ +ǎ 1ǎ 9ǎ
>0¹
ǎ "ǎ %ǎ #ǎ *ǎ
[ǎRǎ Iǎ ǎÏǎjǎŗ ǎj-ǎ FǎRǎ!ǎIǎ\ ǎ I!ǎÏǎ\ ¥ùǎ[ǎ hǎ\ Rǎ!ǎ- irǎeǎrǎǎǎrǎsǎ ǎrǎeǎ ǎǎh}ǎ 'ǎ 3ǎ +ǎ 1ǎ 9ǎ
x¹
#ǎ *ǎ ¥ǎ > ǎ
ǎ #ǎ > aǎ %ǎ
>rǎǎǎ ǎ >4((: 'ǎ 3ǎ +ǎ 1ǎ 9ǎ
ǎ a ǎ " ǎ # ǎ ¥ ǎ
W¹
&(*- +)*!&%)>>¹
G ǎǎ ǎǎ āŭǎ!ǎRǎ!ǎǎ ǎ ǎjJǎǎ!ǎ B
>O¹
G ǎǎs yǎ Tyǎǎ ǎ& ""ǎ !ǎ *ǎ ǎ
>Z¹
G ǎ Tǎ Tǎ ǎ ǎ ǎ @ǎ ǎ ǎ ǎ ǎ Tǎ Tǎ ǎ
G ǎǎ ǎǎ ǎTǎ Tǎ >a¹
#ǎÍ ǎǎ ǎĕǎŀǎǎTƠ ǎ ǎ ǎ#ǎT S ǎ
G ǎǎ ǎ jTǎ yjǎ ǎ y ǎ ǎ ǎT S ǎ ǎ ǎ
fǎ
>f¹
'ǎǎ ǎǎ yŞǎǎS ǎ ǎ ǎǎJǎǎǎ ǎƵyǎ ǎǎ ǎ j ǎ ǎ ǎ ǎcǎǎS ǎǎǎǎ ǎƛ¯ǎÆƟ ǎ ǎyjǎǎ jTǎT nǎ ǎǎǎ y ǎ fǎ
>l¹
x ǎǎ à!ǎĦǎ^ *ǎ ǎàĦǎ\ *]ǎ ǎǎ Tyǎ IàC ƋǎI
>o¹
9 Tyǎ ǎ Tǎ¡ ǎ(ǎ*
>s¹
ǎ!ǎ*ǎ ǎ(ǎ*ǎ02
G ǎǎyǎǎTTǎǎ ǎ ǎTǎǎ;;;ǎǎǎ jTǎjJǎ % M
>x¹
'ǎ ǎǎǎƦ ǎǎ ǎǎǎ$ ǎǎ$ ǎǎǎ ǎ ĭǎǎ-ǎ Fǎ$ ǎǎíJǎ"ǎ ǎ ǎ íǎƲǎ ǎǎ $ ǎ %Òǎ ǎǎǎJǎ"ǎǎǎǎG ǎǎǎ ǎ.ǎ
ǎǎǎǎǎJǎ ǎǎǎǎǎƧ -ǎ
P0¹
'èǎ Jǎ S ǎ ǎnËǎǎǎǎ %ǎ #ǎ ##ǎ #ǎ * ǎ ǎ Eǎ ǎ ǎ ǎ ǎ ǎ ǎ ǎ ǎ Çd ǎ ǎ ǎ S ǎ ǎ ǎ ǎ Bǎ §ǎǎ ǎ ǎǎǎ S ǎ ǎ#ǎ ǎE]ǎ Çd ǎ ǎ ǎǎ ǎ nǎĞǎnǎ
P>¹
Fǎǎ< GH ¹H .M I \ ǎǎǎ ǎǎǎ ǎ ¹ "# G ǎǎ ǎǎ .
PP¹
G ǎǎ ǎǎǎǎ=ǎǎǎǎǎǎǎ<ǎ ǎǎ-ǎ Rǎ \ ǎ =Iǎ =Rǎ \ ǎ I-ǎ
PZ¹
>ǎǎǎǎ?¹Q¹ \¹,,,¹¹ uu¹ ǎ ǎǎǎǎǎ<ǎǎ&ǎ & -ǎx ǎǎ@Q¹!ǎ|**+ uuQ¹ ^ ǎ ǎǎ ǎǎ@c¹(ǎQc¹(ǎ-Bǎ(ǎu 8
8
<@
DG
8
Pa¹
G ǎǎ ǎǎ
A
Pf¹
D
GtI
8
I J
JK
8
8
<22D<22G
1 1 J L
s
8 <22G
8
<22I <22J
!
<22I
8 " <22J <22K
'ǎǎǎ ǎǎǎǎ² ǎǎǎǎǎ"EEǎ% ǎ>ǎJǎǎ
ǎǎ ǎJǎ*ǎǎ ǎǎŔ =ǎ ǎǎǎǎG ǎǎ ǎǎ $= Pl¹
>ǎ ǎǎǎ ǎ $ ǎǎ ǎ ǎ /¹ ǎ ǎǎYǎǎ $ ^ ǃǎ'ǎ ǎǎ ǎƶYǎǎǎǎ ǎĞǎǎǎ /¹ $ ǎ$ G ǎǎ ǎǎ, $
ǎ
/¹
$
d¹
Oo¹
Os¹
Ox¹
>ǎ ǎ ǎ ǎ ǎ ǎ ǎ ǎ ǎ ǎ ǎǎRǎ ǎ ǎǎ -ǎG ǎǎ ǎǎRǎ [ǎ w ' ǎǎǎǎR@ & "Rǎ?ǎǎ^ ǎǎǎǎǎǎ§ǎw ǎ ǎǎR@ǎ?ǎ"R& ǎ\ ǎ ǎǎ ǎ b' [ǎ=ǎǎǎ ǎǎ ǎY Ććǎ=-ǎ1Çd
§ǎhǎ \âǎǎ?ǎâǎǎ?â"ǎǎ ǎ ?ǎBBǎ?ǎŠǎǎÇd ǎǎ ǎǎ ahǎ Z0¹
Z>¹
ZO¹
ZZ¹
[ǎ2ǎ ǎ6ǎǎǎǎǎR@ǎ?ǎRǎ?ǎ ǎ ǎǎǎ ǎáǎǎǎǎǎ R@ & Rǎ? ǎ ǎ\ fǎ G ǎǎ ǎǎ2 ?ǎ6ǎ?ǎ2C á6C áĺÕ @L
"ǎ Sǎǎ"ǎǎǎ ǎǎ ǎǎǎ ǎǎǎǎ Gƞ ǎǎ ǎǎJǎǎ ǎǎ ǎǎǎǎǎǎǎǎ ǎĝƳǎ¯ǎǎ ǎǎ ǎ Sǎ x ǎǎ=ǎǎǎŕ ǎǎã î ǎǎ *RIǎ ǎRǎ¯ĝƿǎ ǎ Jǎ ǎ ǎ ǎǎ ǎ ǎ ǎ ǎ Eǎ ǎ Jǎ ǎ ǎ=ǎ ǎ ǎ ǎǎ
ǎJǎ}ǎ ¹ ƽǎ$$ \ 6K& ǎYǎ$ \ 2ÿǎYǎ ǎ $ ^ 2ǎYǎǎ2ǎ¯ ǎ6
ǎǎǎǎƆG ǎǎ ǎǎ 2& 6ǎ Za¹
=ǎ@ǎC Eǎ ǎJǎ ǎǎ=ǎǎ ǎĆćǎ=ǎŰűǎ ü ǎǎǎ ǎ =ǎ@ǎŖ ǎ * ¯ǎǎǎǎ}ǎ
Zf¹
[ǎ=ǎǎǎ ǎǎ ǎǎ= @ǎ?ǎ E=ǎ?ǎ"ǎǎǎ ǎ<ÕǎG ǎ ǎǎ=ǎ
i¹
YxrcmTcumcxgc}Ygxm Vdj`Ylkm^{QYta^iYtdZYg{SgyildY\{VQS { O00s¹ Ovjdkn{V^Ztdkj{Vkgvtdkjq {
/
4
8gnsOl-
9
4>?{ \ >2.{/>-<{ \ 4<{ <{ >B{ IgN
4--C8{/ǎ4D{ 4>/8{ Ch
pVO
ilhNqMp
MhgpIWgn RIMphln
%
IgN
'+
sVWMV
ilhNqMO
/-08
8gnsOl-
T 4u
w
u E{ \ -{ 4u
(ǎw
/ǎ-{
u
/E{ ǎ hl
E hl
&ǎIgNw
\4u
\ Ŏ#ǎhl
#ǎ/0w
u E{ ^ -{ Ch
sO
VIrO
u
\
-
8gnsOl-
1 EVO
SZlnp
WgpOUOl
Wn
4//{ IgN
pVO
_Inp
Wn
4--/ { Ch
4//{< //-{/ǎ4--/{ ST <M/ǎ/C-
"
8gnsOl-
' *ǎ5..B{ \ C{ /{5..B{ \ #"4þǎ /{ Rhl
nhbO
WgpOUOln
l0{ CWbW_Il_y
sO
VIrO E5..B{[ C{/ 5..B{/ǎ#"4@ǎ/{ Rhl
nhbO
WgpQUOln
]!
=OgMO
pVO
lObIWgNOl
Wn
4
>
8gnsOl-
1
ǎ 4{ 4{ ǎ ǎ >{ "ǎ "ǎ \
#
8gnsOl-
' @p
Wn
M_OIl
pVIp
ƒ/ǎ/{ Ch
$$ \//{ UV$\4 {=OgMO
$$ MIg
LO
ǎC{ hl
"ǎ*ǎhl >{#ǎhl
k¹>{ hl
*ǎ"ǎhl
C{(ǎ
*
8gnsOl-
3 {4{t{5{ \ "ǎ?ǎ4u{ǎ\ *ǎ4u{ IgN
{4{ "2 *ǎ 4t{Ch
{4{u{ 5 ?ǎ4{ t5{/ǎ/ " *¹ CWgMO
-{ Y 4{ tL{/{-{ Y 4{ u5{Y /{ /0pVO
_OInp
WgpOUOl pVIp
Wn
UlOIpOl
pVIg
4{
"
Wn
& ?L
C!
8gnsOl-
T ©¹ ©¹ ©¹ ©¹ +¹ u
?ǎw
\ E{ 3 Chu4
/{ 3w
/ǎ E 3IgN
/ǎ/{ 3 CcMO
u
w
z{¹ -{ sO VIrO
4{ h ¹ ǎ¹/C{
E
8gnsOl-
1ǎ 4--C{/ǎ48{ 4>/{;hgnWNOl
Fu I \ 4U1{ 4>/{oY{ `w
F \ 4U6{ 4>/{o6{IgN
I ſ/ǎ4U9{ 4>/{o:v{EVOgjm
V V \ ǎIgN
¦§¹¦Q¹¦ \/{nh
pVO
gqbLOl
hR
>=D6G6J2M WgpOUOl 5{ 8{ 8{ nh_qpWhgn
Wn
?T5{ r 8T5{ \ ;-{EhUOpVOl
sWpV
pVO
"ǎihnnWL_O
NWnplWLqpYhgn
hR
pVO
nWUgnn¹
.?]ǎ?]ǎ?ǎ]ǎ .?]ǎ ]ǎ ǎ
/-"
.&]ǎ?] &]ǎ.&]ǎ&] ?]
ǎ<ǎǎ ;-{ ={vǎ/4-{
Zgǎ 3ǎ ǎ ǎ Jǎ ǎ ǎ J7''H{ . ǎ /--%{ ėǎ ǎ ǎ ǎ J7{ /. ǎ /--{ vǎ C/ <{ ?{ J . ǎ/--{vǎ@/ {J . ǎ/--{/ǎ=/ {J H{. ǎ/--{ [ 4/{ ǎJ '{. ǎ/--{/ǎ-/% ǎJ7''H{ JH{. ǎ/--{/ǎ4/%
//%
Zgǎ/ >ǎ ǎǎ.ǎ /{7''H{?4ǎ --C{ = {?4ǎ --C{ ^/%
/4%
'gǎ ;= Gǎ'ē|xē\ǎ2 ?ǎ A44 q
Ĵǎ / ?ǎ,2
?6-ǎ ǎ r/==
2 4 A44 /=={vǎ;=#{
/;%
Z\ǎ4C [ǎ ǎǎ ǎ ǎ [ 6]ǎ ǎ/ǎ6ǎ Fǎ6@ǎvǎC{ PQR 6ǎvǎ4%{ vǎ/=e{/ǎ4C%
/=%
Z\ǎ4= [ǎ oǎvǎ ² ǎǎS ǎǎ ǎOǎvǎ ² ǎǎģS ǎ
>ǎǎǎǎǎ ǎǎǎS ǎǎǎ
ǎǎoǎ| O]ǎ ǎ@- ;@{/ǎ4=%
/>%
Z\ǎ ;4 >ǎSǎ ǎ ǎǎbns{<ǎǎǎ ǎǎ4-{ ǎ.ǎ ǎ= ǎǎǎ XǎǎǎSƻǎǎƗǎǎǎ<]ǎ ǎǎ4 ǎ ǎǎ ǎǎǎJǎǎǎ@={&)*4{[ ;4{ ǎ<
/@
Zgǎ ; Xǎǎv ǎ-ǎǎǎǎǎ<ǎr 7{ /Ar{?ǎA-{/ǎ-%{ Fģ ǎǎ r{/ǎA{ ǎ/{-%{ ǎ Iv -Ivǎ;%{
/A%
Z\ǎ4--C [ǎ r{vǎ ]ǎ ǎ ǎNJ J ǎ XǎR@ǎ/ǎ 4--C{ ?ǎ4--Ar{ MNO r{ 4--C r{ ?ǎ/{ [ -#{ ǎ ǎ J ǎǎr{/ǎ4--C%{
/C%
'\ǎ ; ;/@>{ >ǎ< ǎǎǎ/>{?ǎ;-{?ǎ=>{?ǎ%%%{ {JJ({/ǎ/> 44//{/ǎ;;/@>${
/ E%{
'\ǎ4/@-{ ¹ ǎǎǎǎǎ ]ǎ ǎǎǎǎǎǎǎ oǎ ǎǎ ǎǎ=#{ėǎǎǎǎǎ;{ g q¹
W
$ǎRlhb
nVhi
$ IdN
ǎRlhb
nVhi
Oąǎ #$$ǎ %Òǎ/ǎ $0 WW
"ǎRlhb
nVhi
$ IdN
ǎRlhb
nVhi
Oąǎ $$ǎ(ǎ #$$ǎ %Òǎ/ǎ #$0 WWW
ǎRlhb
nVhi
$ IdN
ǎRlhb
nVhi
O
ǎ #$$ǎ(ǎ#$$ǎ/ǎ$$0 pVO
_hsOnp
ilWMO
Wn
#$ $
8dnsOl-
E @R sO
pI^O
bhNq_h
%0ǎpVO
nOrOd
WdpOUOln
%0ǎ ##0ǎ#0ǎ* 0ǎ0ǎty¹ UWrO
0ǎz{ 0ǎ0ǎ 0 0ǎ
=OdMO
pVO
lObIWdWdU
pVlOO
WdpOUOln
MId
hd_x
LO
##0ǎ* ǎIdN
-
8fnsOl-
$$ EVO
OkqIpWhd
Wn
OkqWrI_Odp
ph
Wǎ& 2ǎ\/ǎ.M°¹ $$ ;InO
lǎ @R
.M Y $$0ǎ pVOd
Wǎǎ& 2ǎ\ 5ǎ.MC $$ǎ VIn
dh
lOI_
lhhp
nWdMO
.MC $$ǎ Y $0 IdN
Wǎǎ& 2ǎWǎ5ǎ.M(ǎ$$ǎVIn
Ip
bhnp
ǎlOI_
lhhpn ;InO
lǎ @R
.M e $$0ǎ pVOd
LhpV
æC 2®ǎ5ǎ. $$ǎ IdN
æC 2®ǎ5ǎ.M(ǎ$$ǎVIn
lOI_
lhhpn
sVWMV
UWrOn
"ǎ 05DH5
;InO
lǎ @R
.M5ǎ$$0ǎ pVOd
\ 2ǎ\ 5ǎ. $$ǎ/ǎ$ǎ VIn
hd_x
ǎ lOI_
lhhp
ǎ520ǎ IdN ǎ ǎ.M(ǎ$$ǎ/ǎ"$ #ǎVIn
ǎlOI_
lhhpo
ǎ/ 2DŽǎ"$ #E & 2ǎW/ =OdMO
.M/ǎ$$
8dnsOl-
Cǎ 8NNWdU
pVO
ǎ OkqIpWhdn
IdN
nWbi_WTWdU
UWrOn
=ǎ(ǎC LM5ǎǎ?OdMO
WR
=ǎ/ǎC0 sO
UOp
$ǎ/ǎǎ sVWMV
Wn
WbihnnWL_O
8dnsOl-
$$ AOp
Pǎ hR pVOb
LO
CǎIdN
=ǎhR pVOb
LO
C ǎEVOd
Pǎ(ǎ=ǎ/ǎǎIdN
&É Pǎ( CƩÉ = /ǎ$ǎ Ch_rWdU
Pǎ/ǎ"0ǎ=ǎ/ǎ#"ǎ =OdMO
CĂ Pǎ(ǎ& Ă =ǎ/ǎ$$-
"
8dnsOl-
$$"
ǎ ǎ C ǎ ǎ IdN
pVO
Ìǎ Ìǎ(ǎ:¹ ÌǎķÌǎ(ǎĸ Ch
pVO
dqbOlIphl
5ǎ J¹ Y¹ M d¹ J--h¹ J--n¹ ǎ J--q¹ N
OdhbcIphl
/ǎY¹ M d¹ h¹ s
V
V
lj U[GOn
5ǎ $$" J--n¹ J--q IL
%
/ / 0 L
0L
0L
8dnsOl-
# EVOlO
IlO
* ǎbq_pWi_On
hR
*0ǎ"%ǎbq_pWi_On
hR
ǎIdN
#ǎbq_pWi_On
hR
**ǎ pVIp
IlO
_Onn pVId
%$$Bǎ Ch
pVOlO
IlO
* ǎ(ǎ"%C #ǎ/ǎ $ǎ dqbLOln
Wd
pVO
nOp
sVWMV
IlO
bq_pWi_On
hR $ǎ ǎ *ǎhl
ǎ=OdMO
ilh
L
L_
I
px
/ǎ¹/ %$ -
v¹
'
ǎC + ǎ ǎ ǎ$ ǎ ǎ $ ǎ < -ǎ F 7ǎ $ < -ǎU ǎ$$ 5ǎ$O{$ Oǎ a $ ǎǎ ǎǎǎ ǎǎ ǎǎ ǎ/¹$ ǎOǎ+7ǎO{$ 5ǎC{Y-
4@$
$
Ǝ ǎ
4A$
'Ùǎ=4 ǎ iǎ [ǎǎU+qǎǎǎ¹/C{ ǎ4ǎ ǎ î 5ǎ42ǎ/C{5ǎ46ǎǎ .2ǎ6 5/{ [+Ýǎǎǎ¹/C{O{lcf!{ qǎ ǎ ǎ lcf 4ǎ5ǎ34-!{ +7ǎ ǎcf /{fO{/{ ǎ .426 ð 4ǎlf{O{l{ a . $ǎ3C{5ǎ4ǎ ` lO{@${U ǎf{5ǎ;{
5"B
4C
'
ǎ-{
d{ ǎǎǎǎǎ< ǎ@ & "ǎ {g{O{-{ ǎd 5 =d{ {g{O{-{ ǎd{ ǎǎǎ ǎ< @ǎ {"& g{5ǎ-{ ǎd 5{ {= d g{5ǎ-${iǎ ǎ4g{5ǎ-{ b dk{5ǎg R-$
4 E${
'gǎ4--A{
8 / 8 8 3 8
j m{n{5ǎ m{
;-#
)))
8 / 85ǎm{ /(
b
j 4{n{ j{ ;{n{ ,{ ={n{ {$ $${ j{ n{n
b
4--A{ h·ǎ { ={4--C{ ${$${ ,{ 4--C{4--C{P{ ,4{4--C{ ,{ ;{4--C{ ,
n{
4{
?L
'ůǎ ;4-@= Xǎǎcf{5ǎgh{53ǎ {c {f{5ǎ4---{ ǎg{ {h{\ 4--C${ Fǎǎ
c {gf{ g{ c hf i5ǎXcf{ c {fg{ g{ 5\Xcf c {fh{ {h5\ Q{ 3{ 4---g{ g{ 73{ 4{ ---h{ {h5{ O{ Cg{ {=--Ci{ 5ǎ ;4-@=#{ :.¹
-
'
ǎ*# ǎ
Å ǎǎ ǎǎc ǎ ǎ9s ǎǎ 7ǎ\ #{ǎ "{ǎ E{ǎ & "{"{ǎ! "{ǎ! {ǎ& {ǎ5 %C & #!#!ǎ & #/ *#
'
ǎ ǎ/ǎǎ Bǎ Fǎǎǎ ǎǎ!ǎ*ǎ!ǎǎ!ǎǎ!ǎǎ!ǎIǎǎ Eǎ!ǎǎ!I ǎ ǎ 4ý-ǎ ǎ
ǎ*ǎ!ǎǎ!I ǎ & ǎ!ǎ ! ǎ ǎǎbI& & ǎ\ ǎ4@ǎ_& »¼½ǎ 4@»¼½ǎǎǎ4@ǎuǎǎF Iǎ/ǎǎ!ǎ ǎb ǎ!ǎǎ/ǎ4Ö-ǎßǎ4Öǎ/ǎ0ǎǎ/ǎǎ I5ǎ-ǎßǎ4ýǎ/ǎ 0ǎǎ\ %ǎ
®¹Áǎ#ǎßǎ4Öǎ\ 0ǎǎ/ǎǎ IÁǎE@L
'
ǎ Å ǎ> ǎc< 7ǎ2Vǎ!ǎ2ǎǎ6V& ǎ2!ǎ ǎ& 62ǎ!ǎ ǎ! 6ǎǎF ǎ2! ǎ!ǎ6 ǎ ǎǎ¾¿À2 !ǎ ǎ& 6ǎǎǎ¾¿À2 C 6ǎ·¹ -ǎÅ ǎ> ǎc< 7ǎ ǎ2ǎ!ǎ6V& ǎǎ2Vǎ¾¿À 2C ǎ& 62 & ǎ! 6ǎǎF ǎ2 & ǎ!ǎ6ǎǎǎ¾¿À2 & ǎ& 6ǎǎ ǎ2 & 6 »¼½ǎǎU ǎ2ǎµ 6ǎuǎ
"
Ź
ǎ"# ǎ.=@ǎ& E0ǎ=@ǎ& *ǎ \ ǎ ǎ ǎ.=@ǎ& E0ǎ ǎ\ -ǎ U ǎ=@ǎ Eǎ ǎǎǎ
ìǎ & ǎ ǎǎ=ǎ ǎ ·ǎ F ǎ ǎ "#$M <M»¼½ǎ 0ǎ ǎǎ ǎ !ǎ ǎuǎ"#ǎ ǎ ǎ=
%-
'
ǎ = @ǎ!ǎ E=ǎ!ǎ"ǎ\ .=ǎ!ǎ=ǎ!ǎ #-ǎcǎ=ǎ!ǎǎ \ 4ǎǎǎ ǎ40ǎǎ=ǎ!ǎ #ǎ\ ǎ ǎ #E44ǎ!ǎ ǎ ǎǎǎ<·ǎFǎ ǎǎ 4ǎ!ǎ ǎU ǎ.=ǎ!ǎ =ǎ!ǎ #ĹIJ/ ǎ ǎ=ǎ!ǎǎǎ ǎ.=ǎ!ǎ0ǎ=ǎ!ǎ #ǎ/ǎ ǎ7ǎ ǎǎǎ ǎ<Bǎ [ǎ=ǎ!ǎǎ/ǎP@ǎF ǎ=ǎ!ǎ #ǎ ·¹ Pǎ!ǎ @ǎ\P @ǎ!P ! ǎ ǎ\ =ǎ!ǎǎ! P!ǎ 0ǎǎ PŴǎ#ǎ+ S ǎPǎ \ 0ǎ0ǎ"0ǎ%ǎ ǎ#0ǎ=ǎ!ǎ #ǎ\ P @ǎ!ǎ ǎ\ *0ǎ0ǎE0ǎǎ ǎ"E 7ǎU ǎPǎ5ǎ#ǎ ǎ=ǎ\ ǎ
:;¹
Xwqa
l Satlawfa{ X
fwl
$&%'2(#(!2!*#&2 2 6&&G %( "( %( ( #%'( ( %( (
(
!#%"!( "( #!"#!( 6( "' )
) F &"$") ;&
') ) %) "%") ) (& ) ' )
?' ) &"%)
") 61 )
) ) &% ") ) ')
6+
9k (,.0 A,0. Nu 1 ]t wWS Tnnw nT wWS Ofw]wzRSTrnj0 wn ,. OkR 4]t wWS pn]kw nk wWS t]RS .0 tzQW wWOw 04
L(
.1#' Crn{S wWOw ,1 04L ,4
=Sw noqr PS pnt]w]{S rSOf kzjPSrt tzQW wWOw qrL6+ Crn{S wWOw wWSrS ]t Ok ]kwSVSr tzQW wWOw noIJ :IJ n
>L
qo r&
9k wWS qzORr]fOwSrOf DEFG , . 0 OkR1 OrS wWS j^Rpn]kwt nT wWS t]RSt DE EF FG OkR GD rStpSQw]{Sf OkR BC¹ ]t wWS j]Rpn]kw nT 01 HzppntS [ ]t wWS pn]kw nk wWS f]kS ,hN& tzQW wWOw [0L .0 Crn{S wWOw fQ[}7L u
H]~ R]tw]kQw pnt]w]{S ]kwSVSrt noqrC} _{ OrS V]{Sk ;OQd OkR ;]ff QOfQzfOwSR wWS tzjt nT SOQW pO]r nT wWStS kzjPSrt ;OQd QfO]jt wWOw WS WOt %$ pr]jS kzjPSrt |W]fS ;]ff QfO]jt wWOw tWS WOt !M pr]jS kzjPSrt OjnkV wWS tzjt MWn WOt wWS QnrrSQw QfO]j+
2SwSrj]kS Off pr]jSt
tzQW wWOw
mk @ /
C
]t O pSrTSQw tqzOrS ]S wWS tqzOrS nT Ok ]kwSVSr
6J¹
Xwqa
l Satlawfa{ X
fwl
$&%'2)#)!2!*#&2 2 6&&G %( "( %( ( %"!(
?Su PTL WTL+ IWSk SX
4< OkR PQKp+ - t]j]fOr ur]OkVfSt |S Oftn WO{S SPLoL "{ >Su Z PS uWS on]ku nk PY tn uWOu YZ SY+ 7rnj PSY
=L
>L
|S WO{S
PY:LSP: SY:
¡'' ¹
PZ 4<:Lp :4@L :4@+
IWSrSTnrS
PZPZL¹ / 7rnj WSrS ]u Tnffn|t uWOu PZL OkR |S OrS RnkS
L) MS kSSR un orn{S uWOu Z 9 |}¹ uW]t ]t Sqz]{OfSku un
po r1
- tqzOr]kV PnuW t]RSt OkR t]jof]T
[AB osnr4<+ H]kQS os nr4< U¹ L Z uWS ornnT
]t QnjofSuS
7]rtu nPtSr{S uWOu PRST ]t O oOrOffSfnVrOj P KOr]Vknkt uWSnrSj ?Su uWS S~uSkt]nkt nT Pg OkR RS jSSu Ou h1 H]kQS PT ]t oOrOffSf un
Sh 2PTL eghS$ H]kQS ePgTL ehgS OkR 8TL ^vS gPT OkR ~whS OrS QnkVrzSku tn uWOu ShL TPL SRL [S0 IWzt [ f]St ¸¹ uWS Q]rQfS QSlurSR Ou S |]uW R]OjSuSr Qh& 8SkQS eQ[gMH5 &
_$¹ HzoontS xnTuWS j¹ kzjPSrt OrS S{Sk H]kQS uWS tzj nTu|n S{Sk nr u|n nRR kzjPSrt OrS
< uWS nkf S{Sk or]jS |S tSS uWOu uWS kzjPSr nT or]jSt OjnkV uXS tzjt ]t Ou jntu xD x+ - QWSQd]kV Tnr xL 56 D |S tSS uWOu uWS jO~]jzj {OfzS nT xD x ]t !M OuuO]kSR |WSk xL X& IWzt ;OQdt Okt|Sr ]t RSU_k]uSf |rnkV <]fft Okt|Sr ]t QnrrSQu PSQOztS !M or]jSt QOk PS nPuO]kSR Trnj uWS Tnffn|]kV j¹ kzjPSrt) E¹ ^¹ r ¹ M M ! S{Sk OkR uWS tzj nT u|n R]tu]kQu ont]u]{S ont]u]{S ]kuSVSrt ]t d
M
=Su Bj ¹ >
HL
?
9L
:$ IYSk
IWzt
9
:
;L
BF '9
: Z ¨
5fZjZkOuZkV }S VSu >
: Z
BGB" 6%
}WSrS -¹ ¶¹ W OkR '
:L
9T d -¹ uWSk B \ >
:, IWzt
Z
B OkR uWS
OkR }S WO{S Bj >
: 6$ MS tWOff orn{S P ZkRzQuZnk uWOu By d >x:¹ 6 Tnr S{Sr ZkuSVSr x 9$ IWS ZkSqzOfZu QSruOZkf WnfRt Tnr x 9+ Hn }S OttzjS uWOu Zu WnfRt Tnr tnjS x 9% BnuS uWOu
VZ{Sk S~orSttZnk Zt ZkRSSR O tqzOrS 9T Z - ¹ uWSk
;L
;L
;L
>x¹ 6O'6 >x:¹ 6
:L
>x: 6 lx > W 6 6'6 W B ¹ ' >x:¹ 6 >x:'6 >x:¹ 6
IWSrSTnrS
¨¢8¹
;L
B . ¨¢¹ d B>x:¹ 6d >x'6:¹ 6/
M
Tnre 9-
Wvp`kR`sk~`ve`z Wevk !%'&(- * $*!#-#,$'!- - 4$$E{ %!&(-*!&%-
Wu]p[Xx{5C{ PXx{5%%F{
%I;% .5%%{bp{
$'&(*%*-
$*.)0 0 :>{ (,*+&$* $+)0 /&,)0 $*.)*0&$0+0 $*.)0 *+0')&-0 &)0 +0 #,"+'"0 &0 (,*+&$*0 $+)0 /&,)0 $*.)0 &$0 +0 $*.)0 *+0 /0 *$0+0 ,"0 &$+$$0+0 "++)0 0 0 L{ N{ &)0 0 &))*'&$$0 +& +0 &))+0$*.)0 &)0 +0 &+)0 *&)+0 (,*+&$*0 .)+0 /&,)0 $*.)0 &$0 +0 $*.)0 *+0 $0 *0 +0'')&')+0 ,"0 "&.0 /&,)0 $*.)0 &0*+'*0)0 $0+&0 ,*+0/&,)0 %*.)* 0 0(,*+&$0 ))*0 -{ #)! &0","+&)*0 )0""&. 0
- --- - - - - ----
<e¹
Qwgtdlg^{MakdZ^{Uw^qtdkjq{
ĵǎ ǎ(ǎǎ(ǎDǎ(ǎ*ǎ(ǎ ǎ(ǎ;;ǎ 7¹ G ǎǎǎ ǎǎǎ ǎ ǎ ( M(ǎ"ǎ!# ǎ ǎ £Bǎ! $$ǎ ǎ ǎ! "ǎ .'ǎ ";ǎ ";ǎ 3¡ǎ D$ǎ D$ǎ .+ǎ D ǎ ;ǎ Ó1ǎ ;;ǎ $$ǎ 9ǎ $ ǎ
Fǎǎǎ ǎ LM ǎǎŽǎǎ Ƙǎǎǎ ǎ
lǎ C LMAǎ"C_ŵǎ "CǎLMǎ!ǎ M _ Ŷǎ ǎe ®¹ǎ iǎǎǎ ǎ _} .'ǎ _ǎe& ǎ _ǎ ǎ ǎ 3¡ ǎ .+ǎ _ǎ& ǎ Ó1ǎ _ǎe ǎ 9ǎ _ǎ ǎǎ7ǎǎǎ
cǎ 3¡ǎ .+ǎ
ǎ& ¦¦ 0ǎ Æ Ê. ǎ 2?@.Ċċ ǎ Dǎ Ĉĉǎ.Ĉĉõ "ǎ ƀ¦Ê.¦ǎ ( "ǎ
Ó1ǎ
ŏĊċÊ .¦ǎ
9ǎ
ǎ M õČÊ.ČŐLjǎ
M
M
M
M
"Bǎ
cǎ _ ǎdǎǎ ǎ ǎǎǎǎǎǎǎ< āǎ ?2 ?ǎd\ & ǎǎ ǎǎ _d .'ǎ 6 .3ǎ ñǎ .+ǎ & .1ǎ ǎ .9ǎ 6
Dǎ
'ǎǎ ǎ6&ǎǎǎ ǎ 0ǎǎ6&>ǎǎǎ 7
ǎǎǎBǎq ǎǎ7ǎǎǎǎǎǎǎǎǎ
ǎ7ǎ6 ǎ .'ǎ ǎ ǎ .3ǎ 6 *ǎ .+ǎ ;ǎ ;ǎ .1ǎ ;ǎ ǎ .9ǎ *ǎ
#-ǎ
'3+1ǎǎǎ<ǎǎ'3ǎ] 2]ǎ ǎ'9qxǎǎǎ ǎ ǎƹǎ9ǎǎǎ 3+ǎ ǎ1ǎǎǎ ǎqxBǎ cǎ'9ǎ\ dǎǎǎǎǎ ǎ9q} d 9 .'ǎ 3ǎ +ǎ _ 6_0 .3ǎ d qǎ 2ǎ _1 7 d .+ǎ 6d _1 ! .1ǎ L d 0 _ .9ǎ d xǎ
M
*ǎǎ
ǎ %~ǎ 7¹ ~ǎ *ǎ~ǎ ǎ ~ǎ ďǎ
ǎǎǎ Hǎǎǎǎ ljǎ ő!ǎ Œ ó £ ( ǎ œ!ǎ ǎ ǎ ǎ ǎ .'ǎ ǎ ǎ 3ǎ aǎ *ǎ .+ǎ "ǎ .1ǎ aǎ %ǎ .9ǎ aǎ
.L
.L
.L
'ǎ 8 ǎ ǎ 8ǎǎYǎ8ǎ8 ǎ 8 ǎǎ ǎZ3+ǎǎ©ǎ'+3ǎ ǎ 8ǎǎ cǎ'3ǎ^ ;ǎYǎ ǎǎǎǎǎ8ǎ'3+ǎ8ǎ#ǎ @ǎ` ǎ Hǎ3 .'ǎ aBaǎ 3ǎ a# .+ǎ .1ǎ " .9ǎ ŝǎ
;ǎ
Uǎ7ǎ Hǎ8ǎ=ǎǎǎ ǎǎ *=ǎ!ǎ ǎ ǎǎ ǎ<ǎ ǎ =ǎ(ǎ ǎbǎa$$}ǎ .'ǎ #ǎ 3ǎ ;ǎ .+ǎ aǎ .1ǎ %ǎ .9ǎ ǎ
$
ď ǎǎ8 ǎHǎǎ.!ǎ.
(858 48#8ǎ ǎ ǎHǎǎ- 8H
ƫǎǎ .' 3ǎ #ǎ .+ǎ ;ǎ aǎ .1ǎ %ǎ .9ǎ
M
¹
.M
Vakmt{Uv^qtdkjq{
-
q ǎǎǎ ǎ ǎǎ {O& ǎ$ǎe ǎ x Hǎǎǎ Iǎǎ Hǎǎǎ ǎǎǎ?Ɠǎ Éǎ \ %$$ǎ
×ǎ?ǎ%
¹ ^ %$$0ǎ`d ǎǎs ǎHǎ
q ǎǎǎ Hǎ ǎLǎ ǎǎťǎ e = ǎH7ǎ ǎ ²ǎLǎLǎ(ǎ ¹ Lǎ?ǎK ¹ Lǎ(ǎ ]¹ Lǎ?ǎ"ǎǎ
"
>ǎǎǎǎ ǎǎǎ< ǎǎ$$#ǎYǎ$$*ǎYǎ$$ǎYǎ
ǎY-ǎ cǎǎ ǎǎ ǎ`d ǎǎǎ ǎHǎǎ
%
q ǎǎǎǎ Hǎ ǎǎǎ ƙǎƺǎ < 7 ǎ?ǎƾǎ
#
>ǎǎǎ ǎǎǎ Hǎ ǎǎǎ#| ǎ ǎ>
7 ǎ ǎǎǎ>ǎ ǎ ǎ`d ǎǎǎǎǎ ǎǎèSǎ[ǎ_ǎ ǎ 7ǎǎǎ ǎǎ ǎǎǎèSǎ ǎǎ 7 ǎ ǎǎǎ `dǎǎ ǎǎ<ǎ ǎǎ 7ǎǎǎ ǎǎ ǎǎǎSǎ ǎ
7 ǎ ǎǎǎǎǎ`dǎƚǎ q ǎǎǎ ǎLǎ ǎ _ LŸǎ Bǎ ǎ
B
q ǎǎǎǎ7ǎǎ%ǎǎǎǎ ǎǎǎ ǎǎ ǎ# ǎǎ ǎǎǎǎ ǎ ǎ ǎ ǎǎǎ:ǎ fǎ>ǎǎǎ B
E
q ǎǎ ǎǎC þǎ ?ǎ.C Éǎ ǎ H ǎ7ǎ@C ǎ(ǎ
$
Fǎǎ'3+ǎ ǎǎ ǎ ǎ1 ǎǎ ǎǎ ǎÄǎ ǎ'1Áǎ31ǎÁ +1÷ cǎ ¹'+3ǎ5ǎijǎ ǎǎHǎǎB
:w¹
2! 38
ƅǎ
cǎ0ǎIǎ) ǎĩǎ):ǎ, Hǎ :,ǎ, ǎ)ǎ*ǎ!ǎ Iǎ(ǎEǎ^ ¤#¤0ǎ ǎǎH)ǎ $( I( ĩǎ
øǎ
G ǎǎH)ǎǎ
¤øǎ
¹ ǎ`d:ǎ0ǎ'19ǎ ,ǎ)ǎ: )ǎ ǎ ©ǎ'91ǎ\$t0ǎ) ǎ3ǎ) ǎ+ǎ):ǎ ,
) "$Ñǎ)#$tǎ )$Ñǎ ) "$Ţǎ(ǎ)#$Ñǎ(ǎ)$Ñǎ
ǎ, ǎ¹ , ǎ)ǎ3+9ǎ ,ǎ)ǎƴ ):)ǎ: )ǎcǎ'3ǎ ^ "ǎYN0ǎ+1ǎ\ #ǎYN ) ǎ3+ǎ^ǎYN0ǎ`d ǎǎH)ǎǎǎ 9ǎ
'ǎ
3ǎ
¹
"ǎ
F,ǎ)ǎǎ) Iǎ):ǎ, Hǎ :,ǎ, ǎ)ǎǎ!ǎIǎ^ $$ǎ) ǎIǎ),ǎǎ N)s NNǎH)ǎG ǎǎH)ǎǎCIǎ
%ǎ
cǎ , oǎ & !0ǎ`d ǎǎH)ǎǎ#ǎ,ǎ(
o!ǎ#ǎ ,ǎ oǎǎ ¤ǎ
#ǎ
¨ǎLǎǎǎ, Hǎ :ǎ:ǎ ǎǎ,Nǎǎ ,ǎǎ,N),ǎ) :,ǎ ,ǎ"ǎ) ǎ ǎ,Nǎǎ ,ǎǎ):,ǎ) :,ǎ ,ǎ$"BǎG ǎǎH)ǎǎ¹
*ǎ
cǎhǎ\ 2
8L
ǎ&ġĽƪǎ P 0ǎ`d ǎǎH)ǎǎ ǎh 4! & 4 ǎ Q
M.¹
'ǎ) :ǎ:ǎ ǎ:ǎ, Hǎ :,ǎǎǎ ): ăǎ D0ǎ;D0ǎL0 ) ǎ),S ǎ:ǎ M),,ǎǎ NǎǎM),ǎ NNǎNM Mǎǎǎ:ǎN:, ĕǎ,Í ǎN ,:) ǎ Dǎ),ǎ *Dǎ) ǎ N ǎǎM),ǎ NNǎNM Mǎ *D0ǎ;Dǎ) ǎLǎ ,) ǎ>ǎ),:ǎǎ) ǎ),ǎǎ,)Nǎ),ǎǎ :: ),:ǎG ǎǎM),ǎ,, MǎH)Mǎǎ
;
>ǎ`Í:ǎMǎ,,ǎ)ǎ: )Mǎ'3+ǎ:ǎ'3ǎ^ '+ǎ1ǎ) ǎ9ǎ):ǎ ,ǎ , ,ǎÄǎ) ǎ'+0ǎ:, HMJ0ǎ, ǎ)ǎÄǎ^ "13ǎ) ǎ'+ǎ\ "'9-ǎcǎǎ):)ǎ ǎ<) : M):)Mǎ3+91ǎ ,ǎDǎ N@ǎ) ǎǎ):)ǎǎǎ: )Mǎ
¹ ,ǎRǎ N@0ǎ` ǎRf 'ǎ
$
>ǎ`:ǎMǎ,,ǎǎ : M,ǎ ǎ :,ǎ'ǎ) ǎ3ǎ) ǎ)ǎM ǎ[ǎ ǎ ,ǎ) )ǎǎǎ : M,ǎ)ǎ«ǎ) ǎǎF,ǎ)ǎ«ǎ^ "$ǎYN0ǎ'3ǎ^ " ǎYNǎ) ǎ ):)ǎǎǎ<) : M):)MǎÄ«ǎ ,ǎ$$ǎ N@ǎcǎ¹ ) ǎ6ǎ ǎǎ):),ǎ
: M,ǎ ǎ :ǎ'ǎ) ǎ :ǎ3ǎ:, HMJ0ǎ` ǎǎH)Mǎ ¹
8 8 8
G ǎǎN)s NNǎH)Mǎǎǎ,n Rǎ?ǎ ?%ǎ ǎ, Rǎ?ǎ :)MǎN:,ǎ I5¹
:ǎRǎ:),ǎH:ǎ)MM
£
G ǎǎǎǎ | ǎ ǎ ǎ ǎǎǎ ǎÈǎǎǎ ǎ| ǎGǎsǎ "%#*EEE0ǎ%%%%%%%%%%%0ǎ"%%%*E
G ǎǎǎ ǎ ǎ ǎǎ B
"f
¨h uǎ²ǎ 0ǎ0ǎ0ǎÕf ǎ0ǎ ǎǎǎǎǎǎƱ ǎ ǎǎ ǎǎǎF ǎLǎ ǎǎǎ ǎ ǎ ǎǎs 7ǎ ǎǎǎh ǎ ǎǎLǎ ǎǎ7ǎǎǎÈǎhǎǎǎǎ ǎǎLǎ
ǎ -ǎG ǎǎǎǎǎ ǎǎ
%
[ǎ xo x= xB ǎ ǎǎ< ǎǎǎ ǎ ǎǎǎ 7 ǎǎ*%0ǎǎ x9 bǎx= bǎxB b dž´LJ >ǎÆǎÆ ǎǎǎǎ< ǎl x: 0ǎx= 0ǎ xB "0ǎ xE *0ǎxH BǎG ǎǎ ǎǎ x>33L/ -
7L
7L
7L
II¹
ň ǎ?ǎ ?ǎ ǎ ǎǎ
YxrcmTcumcxgc}Ygxm Vdj`Ylkm^{QYta^iYtdZYg{SgyildY\{VQS { O00s¹ V^jdkm{V^Ztdkj{Vkgvtdkjq {
gǎ+ F ǎǎǎǎǎ ǎǎǎ ǎ ǎǎǎǎ ǎ$ǎ$%ǎ?*ǎ ǎ$ $;;ǎ ǎ %ǎ Aǎ Aǎ ǎ ǎ$"ǎ$#ǎ$ǎ$$$ ǎ ǎ % ǎ GL GL GL
-
Ůǎ1 F ǎǎ ǎ<ǎC Iǎ^ "ǎ&_ǎ ǎ"& Iǎ/ǎǎ$_ǎǎ 5C %_ I/ǎ C *_£ǎF Â Iǎô %_ÂǎC *_ǎ_Âǎ -
£
gǎ3 Vǎ Aǎǎ >ǎ Vǎ& Vǎ Aǎ µ V ǎ ǎ ǎ [ǎKǎ$ ũ& ǎ& Vǎ(ǎVǎ& ǎ -ǎGǎǎǎǎǎ
ǎ& Vǎ& "VǎC ǎ'ǎ ǎ ǎ ǎµ¹ / ìǎ "ǎ X ǎC AûÛ û A ǎǎ ǎŬ& ǎ Vǎǎ CV ǎ ǎ ǎǎ ǎ %ǎ & ǎǎ ǎ ǎǎ>ǎǎ ǎ ǎǎH¹ ǎ ǎ ǎ ǎµ¹ ǎǎǎµ¹ "ǎ
>ǎKǎ$
"
Vǎǎ ǎ ǎ
gǎ Û & ǎF ǎ X ǎ <ǎVǎ " $6ǎ ǎǎ ǎ*&" ǎ$& ǎ2?ǎ6ǎAǎǎƍ:ǎǎ ǎǎ ǎ *ǎ$2ǎ?6& "ǎ$2ǎ ǎ> ǎ ǎǎ*ǎ(ǎ2ǎïǎ6ǎ/ǎǎ ǎ"ǎ?2 5 ǎ ǎ ǎ ǎ2 5¹ ǎ ǎ6ǎ^ ǎªǎ2(ǎ6ǎ/C ǎ
8L
%-
gǎ3 [ǎD
Ðǎ ǎ"ǎ*ǎ ǎǎǎ ǎ D!
Ðǎǎ%ǎǎ- - ǎǎ;ǎ ǎ D%
^ Ðǎǎ#ǎ;ǎ ǎǎǎǎX ǎ ǎǎǎǎǎǎ ǎǎXǎǎ7ǎǎ ǎǎ2 ǎ6 ǎ Ðǎ ǎǎǎ öǎǎǎ;Ŀǎǎ ǎãǎǎ $ 6ǎãǎ ǎ7ǎǎ ǎ ǎ7ǎ ǎ ǎƝǎ ǎ Ą ǎ ǎǎǎ2ǎ ǎ6ǎǎǎ4¹ ǎD!
7č ǎ ǎǎ2ǎ ǎ6ǎǎǎV#¹ $ $ #$ "%ǎ$ ǎ >ǎǎ<ƥ ǎ 7ǎ/ǎ Kǎ # $ $ Aǎ Aǎ %;ǎ ǎ ?L
@L
IW¹
#f
źlǎ9 Xǎ ǎǎ31ǎuǎǎǎÞ91ǎ / ǎǎ9Gfǎ ªǎ @ǎ x /ǎǎǎ<ǎ'Ż+1 u ǎ Þ31ǎ /ǎǎ ǎǎƇ9Gǎ uǎǎǎ Tǎ'9Gxfǎ
9ǎ
3ǎ
BL CL BL
CL BL CL
x
6ǎ
BL DL BL CL L L 2L L CL L L L BL CL BL CL
Gǎ
0{
x
U ǎ9GǎAò-ǎ 6ǎ xǎ *
'lǎ3 ~ǎ %~ǎ ' ' *ǎm
' m Ģçǎ'ʼn( ơçǎ Ŋ( Ƣƣǎ'ŋ( Ģçǎ ó ǎ ǎ ǎ ǎ Aǎ nǎ' ~ǎ( ǎ' ~ nǎ ~ǎ( ǎ ~ ǎ įǎ ǎ ǎ +L
-
+L
8 fǎ ǎ Ĥǎ 8(ǎĎĔǎ
8 m
6L
6L
m
-
m ǎĤǎ0{9 ĎĔǎ0{8 ,L
8
'lǎ+
Xǎǎǎǎ3+ǎ/ǎ'3ǎ(ǎ@ǎªǎǎ ǎ#ǎ/ǎ;ǎïǎ@ǎ ÿǎ(ǎE¢#ǎ ǎFM ǎǎ< ǎ ǎǎ/ǎǎǎǎ/ǎ3G¹ ǎǎ ǎ ǎ f
Ef
'lǎ9 G ǎǎǎ=ǎ(ǎ ǎbǎǎ T ǎǎ=ǎbǎ##Ef h0{ 3
F ǎ*=ǎ(ǎ ǎ ǎǎ ǎ<ǎǎTǎ*=ǎ(ǎ ǎ/ǎh1!{ ªǎ=ǎAǎ fǎªǎ*ǎ ǎǎ ǎǎ h 0{ 3
6L
¹ ǎ ǎ *ǎ ~ ~(ǎ ǎF ǎ*ǎ ǎǎ ǎ
M`¹
ǎǎǎ*ǎǎǎ ǎǎP¢ ǎǎPǎ!ǎ čǎǎǎPǎ/ǎ*4¢ ǎǎ*4ǎ!ǎ ǎǎ ǎǎ4-ǎ PKǎ& ǎ "Eƕǎ& "4 ǎ §ǎPǎ5ǎ*4ô ǎ ǎ ' Aǎ Aǎ *4ǎKǎ&4 ǎ'ǎ ' bǎ##Eǎ¹ Ɛ *ǎ *ǎ *j 4ǎY ##Eǎ ǎǎǎ4/ǎ ǎǎ- - - ǎǎE Ķ ǎ ǎ İǎPǎ5ǎ*ǎ4ǎ!ǎ ǎǎǎǎǎ AǎPKǎ¢ ǎA "E4Kǎ! "4ǎ *4Kǎ!ǎ4ǎbǎ##Eǎǎǎǎ F¹¹ǎǎ7ǎ¹ǎ *ǎ *ǎ 45ǎ ǎǎǎ - - - ǎǎE-ǎ U ǎǎǎ ǎ ǎ ǎǎ
-ǎ
'lǎ+ǎ ǎ " 2!6ǎ ! 2ǎ 6ǎ
8
6ǎ "2ǎ 2ǎ 6ǎ
8
8L
6Kǎ! "2Kǎ 6Kǎ! "2Kǎ -ǎ>ǎǎ ¨ǎPǎ 26ǎ 26ǎ
ǎ "2Kǎ&P62!ǎ6 Kǎ - ǎ>ǎǎǎǎ7ǎ ǎ ǎǎPǎ ǎ6ǎ 8L
PK6KǎÔ #6KǎÃǎ - ǎ>ǎǎ ǎPKǎ& #Ãǎ ǎǎP Ã ǎ" - ǎ
(8+8!ǎ8 %ǎ!Pǎ ǎÃǎEǎU ǎǎǎ ǎǎE-ǎ
>ǎ2ǎ!6ǎ ǎ
'\ǎ F ǎǎ<7ǎ "& ǎǎ' ǎǎǎ ! ǎ AU Û U! ' ıǎ ǎ Ô U 'ǎ ' ǎǎǎ ǎ ǎbǎ#bǎ ǎǎ ǎ ' T¹ -
-
'Ąǎ % ǎ IKǎAǎ!ǎIĘ!ǎI Kǎ& Ĩę ºǎ!ǎIºǎ Úǎ!ǎIKǎ¢ Ĩ!ǎ %Ę%&%ęǎAǎ%ǎ 8L
'lǎ % Åǎǎ ǎ åǎǎǎLǎ/ǎ%-
"
'lǎ # 37ǎǎ<T7ǎǎbǎ#ǎ!ǎ*ǎ!ǎǎ/ǎ# -ǎF ǎ ǎǎǎ ǎǎǎǎǎ ǎǎ ǎ#-
%-
'\ǎ % iǎ
ǎ ǎ ǎ ǎ!ǎǎ & bǎ& ǎ ǎ
Y
Ne¹
)@{
8{
){
){
ºǎǎ
8{
X X
8{
ºǎǎX8{ )@{ B{ 89,+ X q X * X
'ǎ){X89,*A{X ǎǎǎǎ ǎ ǎǎǎǎ < 7ǎǎǎ/ǎ 8{ ǎǎ#ǎU ǎǎǎ%ǎǎBǎ #
'\ǎ E _ǎ/ǎƯĜǎǎåǎWǎ 7 ǎǎ /ǎ"ú #ǎ
ǎ5ǎưĜǎǎåǎWǎ 7 ǎǎǎ ǎ5ǎ ú "B _ǎ > ŲųǎB ǎ ǎ ǎǎǎLǎ/ǎEBǎ Eǎ ǎ 5L
)@!
Z\ǎ # iǎ ŘĀǎ ǎtǎ!ǎŚǎ ǎ"tǎ!ǎ¹ ǎt Aǎş ǎ³ǎ´ ǎ"tǎ´ Ǎǎǂ
0 8 ŤŨ 8 & * tǎ!0 "tǎ ³ 8 řĀ & * ǎ³ǎ!' ǎ#tǎ! ǎt8
AŜƂŧ tǎō tǎ! "³ 8L
5L
Aǎ¹ \]AC Aǎ ǎ ƄǎK ǎ Kǎ U ǎǎǎ ǎ ǎKǎAǎ#ǎ B
'\ǎ 8{# >ǎ7ǎǎ ǎǎ ǎǎǎǎBǎXǎǎǎǎ7ǎ ǎǎǎ ǎǎ ǎ#& ) {/ǎ ǎXǎǎ ǎǎǎǎ
ǎ ǎǎ ǎǎ8{ ǎǎǎǎ ǎǎ>ǎ ǎǎ7ǎǎ ǎǎǎ ǎǎ ǎ8{ ǎ5ǎ#
E
'\ǎ ) 37ǎǎ ǎ ǎǎ ǎ)+##{!ǎ& 30##{ AǎKǎ& ǎ!ǎ ǎļě!ǎ6 ǎǎ ǎǎ< ǎ ǎ2(ǎ6ǎ ǎǎ ǎ ǎ ǎ2ǎ ǎ6-ǎ XǎǎKǎµ ǎ(ǎǎAǎǎ& &ǎ>ǎ ǎ)+##{!ǎ& 32##{ Aǎǎ& &ǎ! ě!ǎ6
Ih¹
8ǎǎ\ ǎǎj8ǎ ð V¸¸ǎA2(ǎ6ǎ ǎ2(ǎ6ǎ\ 8ǎǎ\ ǎǎj8ǎ ǎ& ţ¸¸ǎAǎ2ǎ(ǎ6ǎ ǎ2ǎ(ǎ6ǎ\ · F 8ǎǎǎ<8,ǎ8 ,ǎ2ǎ\ ǎ ǎ6ǎ\ -ǎ
',\ǎ ; h
-
G8,ǎǎǎ,8 ǎī1'+ǎ ǎī13+ǎǎ8,, ,ǎ8,ǎǎ ǎ m1'+ǎ m1+'ǎ ǎm13+ǎ/ǎm1+3-ǎXǎj7ǎ ,8 8ǎǎ,ǎǎ,ǎ8ǎ ë+ ǎ m1'+ǎ(ǎm13+ǎ(ǎm1+'ǎ(ǎm1+3ǎ/ Īǎ m1+'ǎ(ǎm1+3ǎ\ Ī-ǎ >ǎ©'+3ǎ\ ©1+'ǎ(ǎ©1+3ǎ\ ;ǁǎ
',Ùǎ F8 ǎǎzǎ ǎǎǎ,88 ǎǎ *ǎ(ǎzǎ(ǎǎbǎ*ǎ(ǎzǎ(ǎ;ǎbǎ;ǎ(ǎzǎ(ǎ êǎ >ǎ8,ǎ *ǎ(ǎzǎ(ǎǎbǎ#ǎbǎ;ǎzǎ( % >ǎ8ǎ,ǎ ( z( b ǎ -"ǎ ǎ(z(Âǎ -%-ǎ F8 ǎǎzǎ ǎǎǎ8,ǎǎj8ǎǎ(ǎzǎ(ǎǎ\ -ǎU ǎƊǎ(ǎzǎ(ǎǎ\ -
-
Z,Ùǎ iǎ,ǎǎǎ7ǎ8ǎë+ǎ8,ǎ | ǎƸ pǎ´ǎĚ Oǎ Üǎ ǎ ǎp (ǎ Oǎ(ǎÜǎ iǎ ǎ ĚƮǎp(ǎ Oǎ(ǎÜǎ Aǎp(ǎ Oǎ(ǎ ³& p(ǎOǎ Úǎp(ǎ OC pǎ( Oǎ ǎp(ǎ Oǎ Úǎpǎ( ǎOǎ C p Oǎ ǎ Aǎp(ǎ O & ñ ǎp´ǎ Oǎ GL
-
Im¹
8 8
8 ,8 8
1pId5pId9
pIg5pId9 pId5
pId9
pId
5
pId
9 pId5pId9
pId5pId9
2pId
5
pIg9
pId*z5
9
2pId
6
pId9
pId:
nh
pVO
lOnq_p
Rh__hsn
5L
#
7ensOl,
*
9ǎ
wǎ
1ǎ
¹
HO
VIrO
m'w9ǎ[ *{){
\ |
m9+1ǎ IdN
m'9wǎ *}.mw'9ǎ
~
m+19 EVOlORhlO
Ē'w9ǎWn
nWbW_Il
ph
Ē9ż1ǎIdN
Wp
Rh__hsn
pVIp
'wg w9ǎ\ 9+g+1=OdMO
w+@ǎ[.w9.9+ǎ\.'w.+1ǎ\ )&
Mb!
nh
w+ǎ[*
Pb @L
@L
&
@L
7ensOl-
' CWdMO
ǎ
w
*
DžIǎVIn
pO
bItWbqb
rI_qO
WR
IdN
hd_x
WR *
1
&
BhpO
pVIp
v
VIn
pVO
bItWbqb
rI_qO
WR
IdN
hd_x
WR
v ǎ1
w
1
NhOn
EVqn
Cw
\ &
'
\'
@L
'
7fnsOl.
& FnWdU
pVO
WNOdpWpx
J$
K
$
2J
K J JK
K
sWpV
J1
n\
5
IdN K
2
Mhn
5
sO
VIrO nXf
(
5
Mhn(
53
nWd
5
Mhn
5nXd
5
Mhn
5 #nXf
5Mhn
5
3
#
nWd
5M
hn
5
2
nXd
5 &
2
Mhn
5
1
2
8 /,8
>OdMO
) nXf
(
5
) Mhn(
5
&
@L
Iq¹
#
*-
Z\ǎ % Xǎǎ ǎ ǎ 7ǎǎǎ ǎǎLǎǎ ǎ ǎǎǎǎ±ǎ L ǎǎ¹ +<7ǎL ǎŇ ǎǎǎǎ ǎǎ¹ >ǎ ǎ Lǎ Aǎ"-ǎ F ǎǎ< ǎ ǎLǎ % ÷ǎ ǎǎ< ǎ Lǎ ǎ
- --"- $%%--
@L
Z\ǎ Ė ǎ
EL
N
$
ǎ ǎ ǎ C A ǎ ǎ JK U ǎ Ėǎ5ǎ ǎ C
Zgǎ *% iǎSǎǎǎ ǎ ǀ ǎǎ %ǎ;%ǎ ǎ *%ǎiǎ %ǎ[ @ǎ %Øǎ ;%ǎ @ǎ %@ǎ Ɓǎ *%ǎ5ǎǎ %@ǎ F ǎ [+Ý@ǎ %Øǎ@ǎ %@ǎ ǎR{5ǎ[+Ýǎ %@ǎ ǎ@ǎ %@ǎ ǎR{ ǎǎǎǎǎ ǎ ǎǎLǎ ǎ%Øǎ ǎ *%
@L
;
Zlǎ
@L
pǎ
Oǎ
Ɣǎ
+ ǎ1ǎ ǎĠ+1ǎF ǎ+ǎ Zǎǎǎǎ ǎǎĠ+1ǎ[ ǎÎǎǎǎÞ1ǎ ǎ @ F 7ǎǎ ǎ ǎ ǎǎƈ+1ǎ5ǎ& Îǎǎǎ+1-ǎ ǎ Xǎǎǎ+1ǎ5 ǎǎäZ1ǎĻǎǎä+1ǎ5ǎ"ǎ @ǎ "ǎ $ ǎǎä3+1ǎ[ Yǎ ǎ "ǎ >ǎǎ ǎǎ< ǎǎ A% ǎ ǎ ǎǎ ǎ ǎ @L
@L
@L
Iv¹
B
Z\ǎ #
[ǎǎ ǎǎǎ ǎ ǎ ǎ'ǎ ǎ3ǎǎǎYǎ IǎYǎ 7ǎ ǎǎ+ǎǎǎ ǎǎǎnǎǎ3ǎ ǎǎ'+ǎ ǎǎǎ«ǎ >ǎ'+ǎ\ "ǎYǎ ǎ3+ǎ\ ǎYBǎF ǎë+ǎ ǎǎ | ǎ ǎ ǎ Ił Vǎ!ǎ" Vǎ " Vǎ 7ǎ7ǎħǎǎ IC5;-ǎ "ǎ ǎ!I ǎ ǎ ǎ ǎ Xǎ ǎǎê ǎ'3«ǎiǎ ǎ ǎ ǎ Ƒ!ǎIǎ\ %Bǎ F ǎǎǎ ǎ< ƌǎǎ ǎǎ\ ǎ I/ǎ BǎU ǎ
-
nj #B f
'\ǎ *
8 8 8 -ǎn8!ǎ ǎ 8!ǎ%ǎn8!ǎ ǎ !ǎ 8 -ǎn8!ǎ ǎ 8!ǎ%ǎƜË8!ǎ ǎ 8 ǎ !ǎ%ǎ ǎ!ǎ )nǎ - ǎŦǎöǎ 8!= 8 %, 8!< 8B ü Vǎ % ǎ U ǎǎs ǎ ǎ ǎ !ǎ + -C & *B ǎn ! ǎ ǎ !%ǎ ǎn ¬!ǎ ¹
B
Z\ǎ *%% iǎǎǎ ǎ ;Ņ ǎ Jǎ< ǎ ǎs 7ǎ ǎ ǎ
ǎ ǎǎ ǎǎ ǎ <ƨ7ǎǎ ǎǎ | ǎ ǎ ǎ ǎ nǎ7\ǎ9 ǎ ǎ ǎǎ ǎ7ǎǎ ǎÈǎ ǎǎ;ǎ ǎǎ ǎǎǎǎ ǎǎǎǎǎǎǎ ħǎ ǎ ǎǎ ǎǎ
BǎGǎsǎ ǎ ǎ ǎǎ*;;;;;;;ǎ
W-¹
'
ǎ ǎ
!-
ǎ ǎǎ' ¹ ǎǎǎ<
Aǎ^DDB
ZËgǎ D ' ' ' ?ǎ' ǎǎ ǎǎǎ G ǎǎǎǎ ǎ ' ' ǎ ' ' ǎ -ǎF ǎ ' ǎǎ ǎ ǎ ǎ ǎ ' ǎ Bǎ¨ǎ ' Aǎ4ǎ ǎ ' ǎ C o F< ǎǎ ǎǎǎ< ǎǎ ǎ ' 5ǎ4@ǎ ǎ ' ' ^ F² > ǎ ĥĥ5ǎ.j C y Doy Xǎǎ ǎǎǎǎ ǎ ǎ ǎ ' ǎ ' ' ' ǎǎǎǎǎ ǎiǎÍǎ ǎ ¹ 4ǎAǎ ' 4ǎAǎ ' ǎ ǎ ǎǎ ¹ ǎ4ǎ ǎ| ǎǎ< ǎ. ǎǎ>ǎ ǎ ¹ 4ǎAǎ ǎ ǎ ǎǎǎ< ǎǎ ' ǎ ¹4ǎ C ' U ǎ ¹ ǎD F¹ 0{ ' Aǎk ?ǎ ' AǎD B ǎ4ǎuǎkEǎ ǎ ' ǎ4
k
Zgǎ # iǎ`dǎ`d ǎǎ ǎǎ ǎƤǎǎhǎǎǎ ǎǎ +Jǎ' ǎǎ ǎǎ¹ Xǎǎ ǎǎ ǎ4ǎ ǎǎǎ ǎǎLǎǎ4ǎ ǎǎ ǎǎL ǎ>ǎǎǎ¹4ǎ¹' ǎ4ǎ ǎ hǎ4ǎǎǎ ǎǎLǎǎ ǎǎ ǎǎǎǎǎǎǎhǎǎ
ǎǎ ǎ4ǎ ǎ4ǎǎǎǎ ǎǎ¹ U ǎǎǎ
' ǎ ǎǎ¹ >ǎ ǎľǎǎ ǎ5ǎǎ kǎ k ^ǎ Dǎ^ ǎ Dǎ ' ^ ǎ '' 5ǎkǎ Dǎǎǎ ǎǎ¹ +<Jǎ ǎǎ±ǎǎǎh ǎǎǎ ǎǎLǎǎ ǎǎ ǎǎ' ǎ ǎ' ǎ ǎLǎǎǎ>ǎǎ ǎǎ #ǎ ǎ ^ǎǎǎJǎ±ǎ ǎǎhǎǎǎǎ ǎǎ¹ U Lǎ5ǎ×ǎ @ǎ D ^ ' ǎ ' uǎ#k#k ǎǎǎǎ ǎǎLǎ ^ ǎ #ǎ ǎ(ǎkǎ^ #B
?L
D
ZËgǎ ^#k ¨ǎ5ǎ"' ǎ¤ǎśǎǎ^kǎ^Dǎǎǎǎǎ ǎÈǎ' ǎ^D-ǎiǎ`dǎ`d ǎ ǎǎ ǎ ǎǎǎ Jǎ ǎǎ^Dǎ5ǎǎ D@ǎ [ o5ǎ" ' lǎ ǎǎ ǎǎ #' ǎOǎ^ "' ' ' ǎăǎDǎ ǎǎ ǎǎ # ' >ǎ oéǎ Oǎ ǎǎǎ ǎ ǎǎǎǎǎ Jǎ ǎǎ^DBǎXǎǎ®oǎ W5ǎDǎWOžǎ uǎ Dǎ
O 5ǎ Wǎ " ' ' ' ǎ l ǎ Dǎ ǎǎ ǎǎ $' Wǎ 5ǎ D Bǎ3Jǎǎ ǎǎ ņs oǎ ®ǎ æ đ W|WoǎƭOWǎ^ DBǎ>ǎǎƬǎǎ ǎ ǎǎ WǎoǎéOWuǎWođǎ O O®ǎ5ǎ^DC Dǎ kǎ> Jǎ ǎǎ^Dǎ ǎWƏǎ ǎ CWǎoéǎ ' ' ^.Aǎ' Akǎ ' ^ǎ' ' ' ' ' A^k Bǎ XǎJǎ ǎ ǎJǎ| ǎ ǎ ǎǎ ǎ <Jǎ ǎ ǎǎ^D4ǎ °ǎǎǎ ǎ4ǎ ǎ°ǎǎ4ǎT¹' ǎ' ¹ °ǎ¹ ^kBǎXǎ ^D4ǎ °ǎ^Dǎ5ǎ ǎ ǎ ǎJǎ ǎ °ǎ^Dǎ^ ' >ǎǎ< ǎ"!#!&'
?L
W6¹
¬ǎǎǎǎƷǎǎ $\ *%4ǎ(ǎ$ ǎ4ǎS¹ ǎ ǎ ǎ%&M5%&M"ǎ§ ǎ4ǎ ǎ7ǎ4ǎ ǎ 4ǎC
7L
&8
ǎ5M ǎǎ ǎǎ ' ǎ ǎ7ǎ"Bǎ>ǎǎ
"8 %ǎ ǎ ^
5M \ BǎU¬
$\ *%ǎ % (ǎ!$ *%ǎ(ǎ "ǎ
@L
WI¹
*#"ǎ
' ^
Xwqa
l Satlawfa{ X
fwl
$&%'2(#(!2!*#& 2 2 6&&G ( "( &( ( "& '( ( &( (
(
!" &"!( "( "!""!( 5
2 #'!) ) F &#$#)
;& ') ) $) #$#) ) (&!) '! ?' ) &#$) !!#) 50 !
) ) &$!#) !) ')
%
WWǎ QS% HzppntS _ OkR a OrS rStpSQuZ{Sf pnZkut
@Su PQST PS O urOpSZzj |ZuW PT
nk uWS tZRSt PQOkR STtzQW uWOu bQPaZ cST_/ Crn{S uWOu cQaPZ cS_T&
D
2SuSrjZkS Off prZjSt ¥¹ tzQW uWOu
Zt O pSrTSQu tqzOrS Z!S uWS tqzOrS nT Ok ZkuSVSr
6ZkR Off TzkQuZnkt t £¹ £¹ tn uWOu Z
t9
;L
t t 't t Tnr Off Uz OkR
ZZ t= 5 Tnr Off V{|
_% IWSrS OrS %% QnjjZuuSSt Zk O QfzP! 5OQW QnjjZuuSS WOt
M jSjPSrt OkR
S{Sr u|n
QnjjZuuSSt WO{S O jSjPSr Zk Qnjjnk HWn| uWOu uWSrS Zt O jSjPSr |Wn PSfnkVt un _ QnjjZuuSSt!
=Su u o{ R¹ 5+ Crn{S uWOu
WW¹
Xwqa
l Satlawfa{ X
fwl
$&%'2(#(!2!*#&2 2 6&&G ( $( %( ( %$!(
PQa` `t QQf`Q IWzt cPTa cP`a Z 6E5 & IWSrSTnrS cQ`a bQSa Z 6E5 bP`a 6E5 bPTa' 6E5 tn wWOw QSa` `t Oftn QQf`Q" 8SkQS cPQa Z bTS` OkR cQaP Z 6E5 cPQa bQPa ' 6E5 cTS` cST`' dS`T.
% 9wt QfSOr wWOw
D!
=Sw =T{ >
FL
? Z :& IWSk
IWzt
9:Z =D{
9 :
=r{
4L
|WSrS 5 ~¹ W OkR '
5 wWSk ={Wǎ>: IWzt * ={ OkR wWS VZ{Sk S~prStt`nk `t `kRSSR O tqzOrS :T Z 5 wWSk Z OkR |S WO{S =T{ Z >: 6$ MS tWOff prn{S P `kRzQw`nk wWOw =e{ d >x: 6 Tnr S{Sr `kwSVSr x 9+ IWS `kSqzOf`w QSrwO`kf WnfRt Tnre* 9' Hn |S OttzjS wWOw `w WnfRt Tnr tnjS x ´¹ 9& BnwS wWOw 3f`j`kOw`kV |S VSw >: Z =D=so
6&
:T d
> >x 6 : 6 >x: 6 Ex ' W 4 # 6 6 W ={ >x: 6 >x: 6 >x: 6 6 IWzt
Tnr x 9%
=e f{ Z ={=e{ d A>x: 6 d >x 6 : 6 )
t at S`wWSr QnktwOkwf 5 nr QnktwOkwf 639( /fSOrf S`wWSr nT wWS QnktwOkw TzkQw`nmt OPn{S tOw`tU`St wWS rSqz`rSjSkwt" /nk{SrtSf tzppntS wWS V`{Sk QnkR`w`nkt WnfR HSwx`kV ' |S WO{S
t9 /OtS %
>L
t;ti'tit 9)
ti Z 5*
IWSk t9 Z 5 Tnr Off k ¤¹ OkR WSkQS t `t QnktwOkwf 5+
/OtS D"¹
ti' { K{ 5&
IWSk
t!9 '
6 { t9 z
Z 5 |S WO{S `k pOrw`QzfOr { tZ t Tnr Off k ¹ IWSrSTnrS
HSww`kV
Tnr Off k ¹
Z
6
{$
8SkQS { Z
649%
9w Tnffn|t wWOw
t9Z t ] t t * 9t't HSww`kV Z 5 |S WO{S 649 ' ti Z 9tt
k ¤¹ t`kQS t ³¹ 5&
Y`¹
* ;t:&
IWzt t
% >L
649 Tnr Oif
_$
6nrj Ok ZkQZRSkQS jOurZ~
P |WSrS uWS rn|t OrS
ZkRS~SR P uWS QnjjZuuSSt OkR uWS
Qnfyjkt OrS ZkRS~SR P uWS jSjPSrt OkR |WSrS uWS [ b Skur !$' Z ' ZT jSjPSr c Zt Zk QnjjZuuSS [ IWSk uWSrS OrS UZ{S %t Zk SOQW rn| OkR Tnr SOQW [ c uWSrS S~Ztut x tyQW uWOu !%' Z $%'
3L
' IWyt uWSrS OrS UZTu UZ{S %t Zk P+ LS kSSR un orn{S uWOu uWSrS Zt ' {{=
Qnfyjk |ZuW Tnyr %t LZuWnyu fntt nT VSkSrOfZu OttyjS uWOu "' Z ' Tnr ¹
9L
/nktZRSr uWS tyPjOurZ~ . TnrjSR P uWS UZstu UZ{S Qnfyjkt OkR uWS fOtu uSk rn|t 3OQW rn| nT . WOt Ou fSOtu nkS % 8SkQS . WOt uSk %t 9T uWSrS Zt O Qnfyjk |ZuW uWrSS gt uWSk P WOt O Qnfyjk |ZuW Tnyr %t OkR |S OrS RnkS 9T lnu uWSk S{Sr Qnfyjk WOt u|n %t OkR uWyt SOQW nT uWS QnrrStonkRZkV Qnfyjkt Zk P WOt uWrSS %t - QnktZRSr Off uWS nuWSr rn|t |S tSS uWOu ZT kn Qnfyjk WOt Tnyr %t uWSk S{Sr Qnfyjk jytu WO{S S~OQuf vWrSS %t -yu uWZt Zt ZjonttZPfS Ot uWSrS OrS UZTu UZ{S %t Zk P Pyu
n==
6Zrtu Tnr Ok rSOf kyjPSr &' |S WO{S
|ZuW SqyOfZu ZT OkR nkf ZT &' Z ' 9k oOruZQyfOr Tnr knkkSVOuZ{S &' ' '&' c ''&' '&'
|ZuW SqyOfZu ZT OkR nkf ZT &' Z ' JWyt
''''' '''' c ' '''#''' 'c{'''# Z ''' ' '''' '' '' '' # R¹ '''''# = ' ' 3qyOf\u WnfRt ZT OkR nkf ZT
' ''' Z '
Wd¹
+ + +
966
$ !$ "$ #$ $ #$ $
$
!"! $ !$ ! !! $ (!
.#
#+". .
;TÂ !(#'#.
"'. ,(". $+"#. . '. $+". #'. "*. . # . '.
" " '. (
$. +. ,(".
$+"%. 0$
. #' #. ". . '.(#',. ,(". &+"#.
. !-)#'. ""#. (
".
T&
. ( '"$. ". +.
5%
v¤Ø¤ÔŹȤé Ø®¤é ÈÛŤÔé ͧé دԤ¤¤¿¤Å¤ÈØé ×ÛפØ×é ͧé Ø®¤éפØé "ııı#ıTTwı#ı -ı §ÍÔé ß®¹® Ø®¤é×ÛÅéͧéØ®¤éخԤ¤é¤¿¤Å¤ÈØ×é¹×é éÅۿرϿ¤éͧéB*
=- ¯¤Ô¤é Ô¤é 5* ×ØÛ£¤ÈØ×é ؽ¹È¬é ÏÔØé ±ÈééÅØ®¤Åر×éÍÅϤرعÍÈ*é k§Ø¤Ôéد¤éÍÅϤرعÍÈ Ø®¤ãé £±×ÍÞ¤Ôé دØé ¤®é ͧé Ø®¤Åé ×ÍÄÞ¤×é ¤áØ¿ãé ı ÏÔÍ¿¤Å×é È£é Èãé ı ͧé د¤Åé ×Í¿Þ¤é Ø ¿¤×Øé 4 ÍÅÅÍÈéÏÔÍ¿¤Å$é®Øé¹×éØ®¤éŹȱÅÛÅéÈÛŤÔéͧé×ØÛ£¤ÈØ×é߯Íé×Í¿Þ¤ééÍÅÅÍÈ ÏÔÍ¿¤Åé ߯¹®é¹×é×Í¿Þ¤£éãé ÅÍ×Øé×ØÛ£¤ÈØ×j B1 xÞ¿ÛؤéØ®¤é ×ÛÅ
0 ()-"+
g5b
ß®¤Ô¤é (0?BAa £¤ÈÍؤ×éØ®¤é¬Ô¤Ø¤×Øé¹Èؤ¬¤Ô鿤××éØ®ÈéÍÔé¤ÒéÛééØÍé? E- v¤Ø¤ÔŹȤéØ®¤éÈÛŤÔéͧéÏÍ׹رޤé¹Èؤ¬¤Ô飱ޱ×ÍÔ×éͧéSSNFTTTTé Ø®ØéÔ¤éÈÍØéØ®¤é£¹Þ±×ÍÔ×éͧ SSNGUVVP* I$
¤ØéÑ e #a ¤é éÏÍ¿ãÈÍŹ¿é߹خéÔ¤¿éͤ§ª¹¹¤ÈØ×é×Û®éØ®Øé§ÍÔé ¿¿éÔ¤¿é?
ıı8!B 4
8? Ëı  8?Bı È£éÑ "a 4
Né4ı
+
yı v¤Ø¤ÔŹȤéØ®¤é Þ¿Û¤é ͧé8B .
K, Èéد¤é ØԺȬ¿¤é 6) rel="nofollow">ı )ı 8x 6)Dı Đ)ı ( ½ı È£é ı ¹×éØ®¤é §ÍÍÚé ͧé Ø®¤é ¿Ø¹ØÛ£¤é§ÔÍÅé ) ÍÈØÍé nq*é ké¹Ô¿¤é ߹خ飹ŤؤÔé Ăı ¹ÈؤÔפØ×éØ®¤éפ¬Å¤ÈØé )ıØé wı y¹È£é ªı )w +
L( Èé Ø®¤é ØԹȬ¿¤é 6)Dı 6ı 8x Né6)ı :ı È£é )ı 8x ı
¤Øé ı ¤é Ø®¤é ÏͱÈØé ÍÈé 6)ı ×Û® Ø®Øé ²6ı đ)yı y±È£é Ąå. +
+
O* ÈéØ®¤éØԱȬ¿¤é 6)Dı Ø®¤é±×¤ØÍÔ×éͧé Ēı È£é ² ı Ť¤ØéØéØ®¤é±È¤ÈØÔ¤é*>ı Ø®¤é¤áؤÈ×¹ÍÈéͧ *ıŤ¤Ø×éØ®¤é±ÔÛűԿ¤éͧéØԱȬ¿¤é 6)ıØé áı
¤Øé Zı ¤éØ®¤é§ÍÍØéͧéØ®¤éϤÔϤȣ¹Û¿Ô §ÔÍÅé ı ÍÈØÍé >ı È£é 8ıéÏ͹ÈØéÍÈéØ®¤é¤áؤÈ×¹ÍÈéͧé ı×Û®éØ®Øé*ı ı 8âı v¤Ø¤ÔŹȤ Ø®¤éÞ¿Û¤éͧé 68ı
*ı Zı
*ı . +
NUÂ
R* Èéé ÍÈÞ¤áé ÒÛ¡Ô°¿Ø¤Õ¿é 6) Dı ³6)ı ē) Dı ³6)ı Ĕ) >ı د¤é ¤áؤÈ×°ÍÈ× Í¥é ı È¡é 6)ı Ť¤Øé Øé >ı È¡é د¤é ¤áؤÈ×°ÍÈ×é Í¥é 6ı È¡é )ı Ť¤Øé Øé n$ v¤Ø¤ÔŰȤ د¤éÞ¿Û¤éÍ¥ 6s ı T 6)s) +
+
54, yÍÔé Èãé ÏÍ×9Ø:Þ¤é ±Èؤ¬¤Õé Cı ¿¤Øé Àz¯Â ¤é د¤é פØé Í¥é °Èؤ¬¤Õ×é ¥ÕÍÅé ı ØÍé Cı ±*¤* é {¯Â 8x ¼#ı5ıı sı5ıı
¤ı Íßé××ÛŤéدØé+0 B ı v¤Ø¤ÔűȤéد¤éÅá°ÅÛÅéÞÀÛ¤éÍ¥é ı×Û¯éØ¯Ø Ø¯¤é¥ÍÀ¿Í߰Ȭé°È¤ÒÛÀ°Øã -B ax Fı ¹ı óôı ı )+ (+
R UB=
RU
¯Í¿¡×é¥ÍÕé ¤¯é ı Á {¯Â ߱دé ĉĊı B hı 55) ~Íßé ÅÈãé ¥ÍÛÕ#¡°¬°Øé ÈÛŤÔ×é ¬Õ¤Ø¤Ôé دÈé H444é Èé ¤é¥ÍÕŤ¡é ¥ÔÍÅé د¤é ¡°¬°Ø×é 4é 5é <
Bé Eé Hé K é Lé O é Ré°¥éÍÈ¿ãéد¤é¡°¬°ØéEéÅãé¤éդϤؤ¡j 5<$ ¯Ô¤¤é¬°ÔÀ×éké oéÈ¡é té È¡éȰȤé Íã×éÔ¤é ØÍé¤é¿°È¤¡éÛÏé°ÈééÔÍß,é
¤Øé ı ¤éد¤éÈÛÅ¤Ô Í¥éßã×éد°×éÈé¤é¡ÍȤ鰥é oéÇÛ×Øé¿°¤é¤Øߤ¤ÈékéÈ¡é té È¡éké oéÅÛ×Øé¤éפÏÕؤ¡éã ¤áØ¿ãé Eé Íã×,é v¤Ø¤ÔŰȤé Ģã:Âċ^ 5B$ v¤Ø¤ÕŰȤé د¤é ÈÛŤÕé Í¥é D#¤¿¤Å¤ÈØé ×ÛפØ×é ¼Rı 5ıÕı
ı Í¥é "ı5ııD é sı5ı
ı ×Û¯é Ø¯Ø ı ı ııı °×é¡°Þ°×±¿¤éãé . )+ )+
5E* y°È¡é¯Íßé ÅÈãé دԤ¤é¡°¬°ØéÈÛǤÕ×é ¿ã°È¬é ¤Øߤ¤Èé 544éÈ¡éRRRé °È¿Û×°Þ¤ é ¯Þ¤éØßÍé È¡ ÍÈÀãé ØßÍéÍÈפÛذޤ顰¬°Ø×é°¡¤ÈØ°¿* 5H, y°È¡éد¤éÅá°ÅÛÅéÈØÛÕ¿éÈÛŤÕé߯±¯éԤ顰ްױ¿¤éãéB4éÈ¡é¯Þ¤é¤áØ¿ãé B4é¡°¥¨¤Ô¤ÈØ ÏÍ×°Ø°Þ¤é ¡±Þ±×ÍÕ×* 5K, v¤Ø¤ÕŰȤéد¤éÈÛŤÕéÍ¥é×éØéÚ¯¤é¤È¡éÍ¥éد¤éÞ¿Û¤éÍ¥éد¤éÏÔÍ¡ÛØéı
ı
ı
Dé
)+ )+ )+
5L)
¤Øé Q^
¤éد¤éͤ¦ª°°¤ÈØéÍ¥é Xk °Èéد¤é¤áÏÈ×°ÍÈéÍ¥é a ı ı ıA__Cı ߯¤Ô¤é ı `a "®¯ı yı y°È¡ د¤éÈÛŤÕéÍ¥é°Èؤ¬¤Õ×é n2x ®¯ı 'x cdé SSé×Û¯éدØé ĪĮı 3
īģį 5O$
¤Øé cı ¤éد¤éͤ¦ª°°¤ÈØéÍ¥é O cı °Èé د¤é ¤áÏÈ×°ÍÈéÍ¥ EaıııEaıı0ıEaıı!ııawoxııNıısı$EaııNN) )+
)+
v¤Ø¤ÔŰȤéد¤éÞ¿Û¤é Í¥é ĤN ·!ğh 5R,
¤Øé Öı ıı5ı,ı ¤é¦°Þ¤éÈÛŤÔ×é×Ø°×¥ã°È¬éد¤é ¥Í¿¿Í߰Ȭé ÍÈ¡°Ø±ÍÈ×Z
ı ı ııı ı ı,ı 8x Cı È¡ ı ı ,ıı,ı,ı ;ı ;,ı ,ıı,ı 8x ßı y°È¡éد¤éÞ¿Û¤éÍ¥é !ı ı ñèııòéı
ıı, <4,
¤Øé A5ıı ( ) )x ¤ééפÒۤȤéÍ¥éÕØ°ÍÈ¿éÈÛŤÕ×é×Û¯éدØé A
v¤Ø¤ÔŰȤé ı
__ì9
NWÂ
+
ı È¡é¥ÍÔé ı B ı
=5, y°È¡éد¤éÈÛŤÕéͧ餱¬¯Ø¡°¬°Øé°Èؤ¬¤Õ×éÍÅÏհװȬéد¤é¤°¬¯Ø顱¬°Ø×é§ÕÍÅé5éØÍéOé×Û¯éØ¯Ø Xıı !a ¡Í¤×éÈÍØé°ÅŤ¡°Ø¤¿ãé §Í¿ÀÍßé  §ÍÕé¿¿é Xı دØéÕÛÈ×é§ÕÍÅé5éØÍéL$ ==*
¤Øé nHtı 8x ħııAOııO ı  !àäݬ§ı ıN O Nı ߯¤Õ¤é ĨRıA5ı Rı!ı È¡é Nı Õ¤é ÍÈ×ØÈØ× ß°Ø¯é Nı a %.ı ¯¤Èé ¡°Þ±¡¤¡é ãé ıv M%%Dı ı M%%?Dı ıv M%%/Cı ıv M%% ı È¡é ı M%%@D Õ¤×Ϥذޤ¿ãé nHtı ¿¤Þ¤×ééդűȡ¤Õéͧé M?Ñı / D / ı È¡é M?ı {±È¡éد¤éÞ¿Û¤éͧé ÁHM%%th =B$ y±È¡éد¤éÈÛŤÕéͧé54¿¤ØؤÕéϤÕÅÛØØ°ÍÈ×éÍÅÏձױȬéEé ×Cı Bé×é Bé ×é×Û¯éدØéÈÍéØßÍ ¡»¤ÈØ鿤ØؤÕ×éդ鱡¤ÈØ°¿$
<E*
¤Øé nHtı 8x O !ııııDı ߯¤Õ¤é ı °×ééÕ¤¿éÈÛŤÕ$é }±Þ¤ÈéدØé د¤é °ÈÞ¤Õפé§ÛÈرÍÈé ͧ n ¤â°×Ø×éÈ¡é°×鬰ޤÈéã YaAı D a
ıı
Dx
+
*
/ 2éı;x
A¥!ı ı
Dx
* *
0 ıv
/ ı$;x
A¥!ı
*
߯¤Õ¤é Iı éÈ¡é ı Õ¤éÏÍױذޤéÍÈ×ØÈØ×é «²È¡éد¤éÞ¿Û¤éͧé ıoB ıoBBı =I$ o¤Øߤ¤Èé5é È¡éO444é °È¿Û×°Þ¤é «²È¡éد¤éÈÛŤÕé Í§é °Èؤ¬¤Õ×é ߯±¯éդ顱ްװ¿¤éãé Ȥ°Ø¯¤Õ 5CéÈÍÕé =5éÛØ顱ްװ¿¤éãé ¤°Ø¯¤ÕéEé ÍÕé K$
@Mé
+ + +
966
$ !$ "$ $ "! $
5$ kÈ×ߤÕ`é M?ID7, yÍÕ·hé 78= ¿¤Øé 3:Hhé "Oı 4H 80 Oı0 8=7 È¡é Oı 7x  eq?
¤ı ¯¤Èé 43:;Hhé D7& § "ı N + °×é é B¤¿¤Å¤ÈØé ×ÛפØé ͧé د¤é ¬°Þ¤Èé פØé د¤Èé Bé ¡°Þ°¡¤×é ıı éı Nx °§é È¡é ÍÈ¿ãé ±§ ¤âØ¿ãé ÍȤé ͧéد¤é§ÍÀ¿Í߰ȬéÍÈ¡°Ø°ÍÈ×é¯ÍÀ¡×[é ° + À¿é ıé Nx Õ¤é °Èé 3 @H ÍÕé °Èé 3!H ÍÕé °Èé \ rel="nofollow"> °° + ÍȤé ͧé د¤é Ïıé Nx °×é °Èé 3 H ÈÍد¤Õé °Èé l7é È¡é د¤é د°Õ¡é ÍÈ¤é °Èé \>, ¯¤é ÈÛŤÕé ͧ B¤¿¤Å¤ÈØé×ÛפØ×éͧé 3:H °×é G H yÍÕ餯é¯Í°¤éͧé ı °Èé 3 H °Èé 39 È¡é Nx °Èé 3 #H ß¤é¬¤Ø é B¤¿¤Å¤ÈØé ×ÛפØé ×Û¯é دØé Bé ¡°Þ±¡¤×é ıı éı N&x ¯Û×é د¤é ØÍØ¿é ÈÛŤÕé ͧé B ¤¿¤Å¤ÈØ hé M?ID7. ×ÛפØ×é "ıé N + ×Û¯éدØé ? ¡°Þ¸¡¤×é ı é ı Nx °×é¤ÒÛ¿éØÍé ? () ıD7@ =, kÈ×ߤÕYé F+ °Ø¯ÍÛØé¿Í××éͧ鬤Ȥտ°Øãé ߤéÅãé××ÛŤéدØé¤Þ¤ÕãéÏÕÍ¿¤Åé°×é×ÍÀÞ¤¡éãé ×ÍŤé×ØÛ¡¤ÈØ(
¤Øé 3H ¤éÍȤéͧéد¤é×ØÛ¡¤ÈØ×$é oãé××ÛÅÏØ°ÍÈ餯éͧéد¤é M Íد¤Õé×ØÛ¡¤ÈØ×é×Í¿Þ¤×éØ鿤×Ø ÍȤéÍÅÅÍÈéÏÕÍ¿¤Åé×é3H oãéد¤éÏ°¬¤ÍȯͿ¤éÏÕ°È°Ï¿¤é د¤Õ¤éÕ¤éØ鿤×ØéBé×ØÛ¡¤ÈØ×éß¯Í ×Í¿Þ¤é é ÍÅÅÍÈéÏÕÍ¿¤Åé ߯°¯é °×é ¿×Íé ×Í¿Þ¤¡é ãé 3H ¯¤Õ¤§ÍÕ¤é د¤é űȰÅÛÅé ÈÛŤÕé ͧ ×ØÛ¡¤ÈØ×é ߯Íé×Í¿Þ¤ééÍÅÅÍÈéÏÕÍ¿¤Åé ߯±¯é°×é×Í¿Þ¤¡éãé ÅÍ×Øé ×ØÛ¡¤ÈØ×é °×éØ鿤×Øé D- § د¤éÅ°È°ÅÛÅé ÈÛŤÕé±×é ¤âØ¿ãé D د¤Èé ¤¯éÏÕÍÀ¤Åé±×é ×Í¿Þ¤¡é ãé ¤âØÀãé D ×ØÛ¡¤ÈØ×$é § د¤Õ¤éÕ¤é bx ÏÕÍÀ¤Å×é°Èé د¤éÍÅϤذذÍÈé د¤Èé Dv ?7+ oÛØéد°×é °×ééÍÈØÕ¡°Ø°ÍÈéװȤ D ¡Í¤×éÈÍØé¡°Þ°¡¤é ?7- ¤È¤éد¤éűȰÅÛÅéÈÛŤÕéͧé×ØÛ¡¤ÈØ×é߯Íé×Í¿Þ¤ééÏÕÍ¿¤Åé߯°¯ °×é ×Í¿Þ¤¡éãé ÅÍ×Øé ×ØÛ¡¤ÈØ×é °×é Øé ¿¤×Øé F+ ¤éׯÀ¿é ׯÍßé دØé F °×é°Èé§Øéد¤éÅ°È°ÅÛÅé°È د¤é§Í¿ÀÍß°È¬é ¤âÅÏ¿¤é ߯¤Õ¤é ߤé ßհؤé 8=? ØÍéŤÈéدØéé ×ØÛ¡¤ÈØé ×Í¿Þ¤×é ÏÕÍ¿¤Å×é 8= È¡é ?S 8=TA 8?D 8DF 8FI 8I= =BF =DF =DI ?DI ?FI +
B* kÈ×ߤÕaé IIIIIÍؤéدØé PQUK87$Aahé 7 È¡é PQVK8S$Bahé #0 °§é =k + Éé 3
=k9 & °È¤é I=?J hé =9>ı=8D8 ߤé¯Þ¤é ÃU É Aahé 8= §ÍÕé =9>0 + 0 I=?J È¡éد¤Õ¤éÕ¤é =8D= ×Û¯é É×$é ¯Û× > 7ı8=># = ı==@# => ı ıı88=9>$ =99 ı 8= =8D= 88 =9># =ı => ı g ıı=99 ı8= =8D= 88 =9># ==9 9 # 8 ı8= =8D= hé IIIII/ =7 =99 ı8= =8D= ı=
C, kÈ×ߤÕ[é MMMMM. Íؤé دØé MMKhé =
DMM È¡é DMM °×é é ÏհŤ$é kÈãé ¡°Þ°×ÍÕé ͧé MMKENNNN ¯×é د¤é §ÍÕÅ ı =bDMMd ߯¤Õ¤é ı È¡é éÕ¤é ÏÍ×±Ø°Þ¤é ±Èؤ¬¤Õ×é ¤Øߤ¤Èé 7 È¡é DMMMM& ¯°×é ¡°Þ±×ÍÕé ¡Í¤×éÈÍØé ¡°Þ°¡¤é MMKENNNL ÍÈ¿ãé °Èé ØßÍé פ×é ߯°¯éÕ¤é ¤°Ø¯¤Õé Ixhé DMMMM ÍÕé é hé DMMMM& È Ø¯¤é «²Õ×Øé פé ı Èé ¤é =ENNNN=ENNNNDMM=ENNNNDMM>0 ( ( =ENNNNDMMENNNN1 Èé د¤é פÍÈ¡ פé ıÈé¤éDMMENNNODMMENNNx=DMMENNPN=>) /) DMMENNNN=ENNNN- È餯éפé د¤Õ¤éÕ¤ I4444é ÏÍ××°¿¤é Þ¿Û¤×é ÛØé د¤é ÈÛŤÕé =ENNNNDMMENNNN ±×é ÍÛÈؤ¡é Øß°¤*é ¯Û×é د¤é ØÍØ¿ ÈÛŤÕéͧéÕ¤ÒÛ°Õ¤¡é¡°Þ°×ÍÕ×é °×é =
F7777# 8 MMMMM& +
+
@Qé
I$ kÈ×ߤÔYé ;33N$
¤Øé 4ı ı 'ıı ı Û×Ø°ØÛذȬé د°×é ±ÈØÍé د¤é ¬°Þ¤Èé ¥ÛÈرÍÈ¿é ¤ÒÛرÍÈé ß¤é ¬¤Ø 'Y'ıYı ı'ı+'ı $ ¥ÍÔé¿¿éÔ¤¿é.ı ±È¤é'ı±×ééÏÍÀãÈÍÅ°¿é ߤéÅÛ×Øé¯Þ¤ 'ı+ 'Yı 5x Iı ߯¤Ô¤é ı°×ééÔ¤¿éÍÈ×ØÈØ$é ¯¤éÏÍÀãÈÍű¿é 8ı 7x 'ıY'+ ÈØé ؽ¤é د¤é ׍é Þ¿Û¤é Øé Èé °È«²È°Ø¤é ÈÛŤÔé Í¥é ¡±×Ø°ÈØé ÏÍ°ÈØ×é ×é °Øé °×é ¤Þ¤ÈØÛÀ¿ã ÅÍÈÍØÍÈ°é ÛÈ¿¤××é°Øé °×éé ÍÈ×ØÈØéÏÍ¿ãÈÍÅ°À$a ¤âØé ¿¤Øé 'ı 6x =ıı ı Û×Ø°ØÛرȬ د±×é±ÈØÍé '+ '+ ıı Iı ߤ鬤Øé =ıı=+ ı ×ÍéدØé =ı 4x Iı ߯¤Ô¤é °×ééÔ¤¿ ÍÈ×ØÈØ$é ¯¤Ô¤¥ÍÔ¤é 4ı ı ı ı ı ı Øé Èé ¤é ¤×±Àãé Þ¤Ô±«±¤¡é دØé Èãé ÏÍÀãÈÍÅ°¿ 4ı}ı ıı ııı ×ذ׫±¤×éد¤é¬°Þ¤Èé¥ÛÈرÍÈé¤ÒÛØ°ÍÈ$é װȬé4Bıı NéÈ¡é 4ıı A; ߤ鬤Øé 4ı~ı ı oB ıN$é ¯Û×éÐD3a ı ;33N$ K$ kÈ×ߤÔZé D$ °È¤é ı °×é顰ŤؤÔé ߤé¯Þ¤é č ı ı S3èé ×ÍéدØé ı ±×éÏÔ¿À¤¿éØÍé
?
k×é Ď ı±×éױŰÀÔéØÍé ď Iı ߤé¯Þ¤é ı 6x ı×ÍéدØé ı ı
£ ±Å±¿Ô¿ãé ı ı ı
ı ¯Û×é ı ă ı ı ı D* L$ kÈ×ߤÔbé 8AJ$ È鬤ȤԿé ߤ鿤Øé ı 6x 5ıı i È¡é ı ı
¤Øé ı ¤éد¤é±×¤ØÍÔé Í¥é $ı װȬ Ø®¤éȬÀ¤"±×¤ØÍÔéد¤ÍÔ¤Åé
ı ı· ı ¯Û×é
ıı ıÈ¡é ı oãéؤßÔØ×éد¤ÍÔ¤Å
ı ı ı
ı
6x
ııı
߯°®é¥Ø¤Ôé×ÍÀޱȬé¥ÍÔé ı ¬°Þ¤×
ı ;ı Y
?
O)Â
ı
ı ı ı #ı ıÌı
Û×Ø°ØÛذˬéØ®¤éÞÀÛ¤×éÍ¥é ıV ıı {#ıNx ı K ߤ鬤Øé ı{ı ¯Û×é W ı °×é°×Íפ¿¤× Ë¡éØ®¤é°×¤ØÍÕé ı Í¥é $ ı °×éϤÕϤˡ±Û¿ÕéØÍé ı Ë¡éı °×éد¤éű¡ÏÍ°ËØéÍ¥é ı k×
}ı>ı ߤé®Þ¤é ıı .ı װˬéãØ®¬ÍÕ×éØ®¤ÍÕ¤Åé ߤ鬤Øé ı ı Ø ı ı {ı +'+ ı ı /?ı ®¤Õ¤¥ÍÕ¤ é ?ı ı/ı
O$ kË×ߤÕYé
¤Ø×éÏÕÍÞ¤éØ®Øé W Z8ı °×é ױŰ¿Õé ØÍé W*ı 8
y°Õ×Øé $* ı ı $ ı ı $±Ā ı k×é $* ı ı $* >ı ß¤é ¯Þ¤é $* ı  $* ı }ı ı 8.ı ®°×éŤË×é W* 8ı $
ı  $* ı ×Íé Ø®Øé $ *āı ı $ *>ı Ø®Û×é *ı °×é é Õ°¬®Øˬ¿¤¡é Øհˬ¿¤é ß°Ø®é $ġı 8ıı ( ½.ı k×é $8ı °×é é ÍÅÅÍËé ˬ¿¤ é ߤéØ®Û×é¯Þ¤ W Z8ı °×é ×°Å°ÀÕé ØÍé W* 8^ı ~¤Ë¤ é 8 Zı ı *8* ı ı * * £ı tÍËפÒÛ¤ËØ¿ã
8ı L
* Zı L
* ı ıh R* kË×ߤÕZé ^ <
C
±Ë¤é $ı °×é Ëé ¤àؤհÍÕé ˬ¿¤é ¥ÍÕé Ø®¤é Øհˬ¿¤é >ı ß¤é ®Þ¤é $ı ı $ ı $ ı ¯¤Ë¤é $ı ı $ ı  $ .ı Øé ¥Í¿¿Íß×é دØé $ ı ı $ >ı ×Í Ø®¤é ØհˬÀ¤×é ı Ë¡é ±ı Õ¤é װű¿Õ-é yÍÕé Ø®°× é ߤé ÍØ°Ëé ı ı ı oÛØé $ ı ı $>ı ×Íé ß¤é ¬¤Øé ı ı .ı ױˬé Ø®¤×¤é ¤ÒÛÀ°Ø±¤×é ß¤é ¬¤Ø ı ı ııEı )+
)+
54$ kË×ߤÕ[é (( y°Õ×Øé××ÛŤéØ®Øé +0 ¿Â ı tÍË×°¡¤ÕéØ®¤é¥Í¿¿Í߱ˬéפØéï
"ı ı[ı«ı $ [ı ı #ı5ı5ı
O3Â
'+ '+ '+
5ı(-Eı
ÍÙ¤éÙ®Ùé&Fı ı%Òı ı #$ $ Ë¡ -B
`x
F <ı kıSı
RUB>
¤Ë¤éÙ®¤éË×ߤÔé±×éËÍÙéÀÔ¬¤ÔéÙ®ËéRR%é Íßé ÍË×±¡¤ÔéÙ®¤éפéÙ®Ùé +0ı ((ı yÍÔé \ııııı
$+ %+ %+
Kı(ı
ĩ ı"H\ Qııbîı÷øı bõöı -Sı ÍÙ¤éÙ®Ùé
Íßé××ÜŤéÙ®Ùéı ±×éËãé×ÜפÙéÍ¥é $ ß±Ù®é&Fı GH %gı oã鱬¤ÍˮͿ¤éԱ˱ÏÀ¤é&3m0ı GH ı ¥ÍÔé×ÍŤé \iı ı ı \ı (ı ¤Ùé ;¦Kı;7ı A
3ımgı ®Ü× -B
`x
& < ùúı0;¦ı ;70ıûüı HH\ QııQ HH\ QııQı kı%Sı
RUB>
R6U
kË×ߤÔ\é
Nf
¥é ?ı ÏϤÔ×é ¤áÙÀãé Ù®Ô¤¤éٱŤ×é Ù®¤ËéÙ®¤éËÜŤÔéÍ¥éßã×éÙÍé®ÍÍפé Kı;Kıı ±×é S ~¤Ë¤éÙ®¤éË×ߤÔé±×
/ı K
Hı(ı K
Né K
@ı
NéıQı kı /Þ
K
5<* kË×ߤÔ\é %?
¤Ùé jı ¤éÙ®¤éפÙé Í¥é ÔÔˬ¤Å¤ËÙ×é Í¥é Ù®¤×¤é ¬±ÔÀ×éË¡é Íã×é ÜË¡¤ÔéÙ®¤é ÍË¡±Ù±ÍËé Ù®Ùé Ôı ÅÜ×Ùé ¤é פÏÔÙ¤¡é ãé ¤áÙÀãé Cé Íã×$é yÍÔÅé é À;é mı ß±Ù®é ıı Ùé Ù®¤é ÙßÍé ¤Ë¡×é Ë¡ ¤áÙ¿ãé ?ı Íã×é¤Ùߤ¤ËéÙ®¤Å/é ®¤éËÜŤÔéÍ¥éßã×éÙÍé¥ÍÔÅé×Ü®ééÀ;é mı ±×
ı J
®¤Ìé ߤé®Þ¤é
Fj&ıkıı J
*
K
?]S
*
K
]ı K
@]
×é ߤé Ëé ÍË×±¡¤Ôé ×Ü®é é À;é mı ×é ÍË¤é ±Ù¤Åé Ë¡é Ù®¤Ô¤é Ô¤é ×Ù±¿Àé /ı Íã×é Ë¡é ÍË¤é ¬±ÔÀ ±.¤%é Êı ÍÙ¤é Ù®Ùé ±ËéËãéÔÔˬ¤Å¤ËÙé ±Ëé jCıı ±×éÍÜÙ×±¡¤é Ù®¤éÀÍ¾é ¤Ùߤ¤Ëé ı Ë¡é Eı Ëé¤áÙ¿ã ®¿¥éÍ¥éÙ®¤éÔÔˬ¤Å¤ËÙ×éÍ¥é jı ı ±×é¤Ùߤ¤Ëéı Ë¡é gı ¤Ë¤ +0 kı&jĈıı
*
¤Ë¤éÙ®¤éË×ߤÔé±×é &+
?]ıı %?.
O:Â
]ı
@]S
5B$ kË×ߤÔ]é ( ¢ yÍÔé [ı ı ııCı À¤Ø ÍؤéØ®Øé
* "* * * !* * "* * *"* * !*"* ı Iı
ı"bıiıı¿ıbıÀı ıb 9 [ı ıÂ $ -ıÝ ı@ı ı@ı Ë¡é
ØéËé¤é×®ÍßËéØ®Øé¥ÍÔé ı ı#ılı"ıı$ Dé
$+ %+ &+
ı -Iı ı ı ıııı ±×顱ޱױÀ¤éãé $ ±¥©
ı ¥ÍÔé¿Àé [ı ııýı ÍÔé
±a<"ı ıı-ı
±±a<"ı ıı-ı3ı ±±±a<"ı ıı-ı
$ ¥ÍÔé ×ÍŤé [iııÀı[ı ¿ıı ÍÔ
ıı Ë¡é º"ı ıı-ı $ Ë¡é
º"ı ıı-ı
ı ¥ÍÔé ¿Àé \ıııı ÍÔé
ı D%é
±Þa<"ı ıı-ı
®Û×éØ®¤éËÛŤÔéÍ¥éD¤À¤Å¤ËØé×ÛפØ×é "ı ıı-ıÍ¥é "ıı$ #ı ııı ±×顱ޱױ¿¤éãé $ ±×é
#+ #+ #+
ı -ı×Û®éØ®Øéı ı ı
& *&* '"*( * '"*(* &"* * * )" * * ı
ı
I
$
ıı
I
@7ıı
ı
ı ( Tı
5E, kË×ߤÔYé ®¤Ô¤éÔ¤éØßÍéÏÍ××±¿¤é¥ÍÔÅØ×é¥ÍÔéخԤ¤é¡±¬±ØéËÛŤÔ×éØÍé®Þ¤éØßÍéË¡éÍË¿ãéØßÍéÍËפç Ûرޤ顱¬±Ø×鱡¤ËرÀY ±a ı ß®¤Ô¤é ı > ı Ë¡é ı > Iı ÍÔ ±±a ı ß®¤Ô¤é ı > ı Ë¡é ı > . ®Û×éØ®¤éËÛŤÔéÍ¥é×Û®é±Ëؤ¬¤Ô×é±×
(ı I
(ıı(ı J
(ıı ¡ 5H$ kË×ߤÔ^é / . ¤Øé ı ¤éé ËØÛÔÀé ËÛŤÔ顱ޱױ¿¤éãé B4$é ®Û×é ı Ëé¤é¤àÏÔ¤×פ¡é×
ß®¤Ô¤é p
>?é Πͥé ı ±×
*
EF $ Ë¡é 7ıþÇıVı Ô¤éÏÍױرޤé±Ëؤ¬¤Ô×$é ®¤éËÛŤÔéÍ¥éÏÍױرޤ顱ޱ×ÍÔ×
%** %**
ı7ıııı!ııııVııı ±Ë¤é ıı =
$
/Iı ß¤é ®Þ¤é p
ı ı ±¥
7ıı!ııVıı
ı ıT
ııı
¤Ë¤é Ĝı Ë¡é ı ®×é ¤àØ¿ãé A3é ÏÍ×±Ø±Þ¤é ¡±Þ±×ÍÔ×é ±¥©é ı ı "$ Ë¡é 7ıı!ı Vıı ıı Cı ß®¤Ô¤é 7ı!#ıVı Ô¤é¿ÀéÏÍױرޤ.é yÛÔØ®¤Ô é 7ıı!ııVıııı ß®¤Ô¤é 7# !Vı Ô¤é¿Àé ÏÍױرޤ é ±¥©é "7!V-ıı"- ®¤éÅà±ÅÛÅéÞ¿Û¤éÍ¥é×Û®é Ix ±×é ıı
7ı
/Nı ı/
OMÂ
5K, kË×ߤÕYé F44&
¤Øé m_x ¤é د¤é Åà°ÅÛÅé °Ëؤ¬¤Ôé ×Û¯é دØé =u °×éé ¥ØÍÕé Í¥é 8
=
?
D
=44K Ë¡é v د¤éÅà°ÅÛÅé°Ëؤ¬¤Õé×Û¯éدØé Fw °×éé¥ØÍÕéÍ¥é 8
=
?
D
=44K& '+ '+ '+
!+ !+ !+
¯¤Ëéد¤éËÛŤÔéÍ¥é×éØéد¤é¤Ë¡éÍ¥é 8
=
?
D
Øé±×éÍÞ±ÍÛ×éدØé qı A v&
"+ "+ "+
=44K
±×é¤ÒÛ¿éØÍé +B
`x
"q5ıv&
£Â
פÕÞ¤é دØé =44K 3
FG' yÍÕé 8 c b D د¤é ËÛŤÕé ¡¤ËÍؤ¡é ãé ai* Í¥é °Ëؤ¬¤Õ×é ±Ë 5ı=44K ¡¤ËÍؤ¡éãé 3H ¡°Þ°×±¿¤éãé Fi ±× "#ı=?D !+ !+ !+
ai 8x p=44K3Fi] (
Íé a9 8x D48 a 8x K4 a > @ ãé FEY
+
8I a
E 8x ?& ¯Û×é د¤Ô¤éÕ¤é¤àØÀãé ? ËÛŤÔ×é°Ëé 3H ¡°Þ±×±¿¤
¤àØ¿ãé 8I" ? 8x 8? ËÛŤÕ×é°Ëé 3H ¡¶Þ°×°¿¤éãé F@ ÛØéËÍØéãé FEZ ¤àØ¿ãé K4# 8I 8x ID ËÛŤÕ×é°Ëé 3H ¡°Þ±×±À¤éãé F> ÛØéËÍØéãé F@Z ¤àØ¿ãé D48# K4
+
?=8 ËÛŤÕ×é°Ëé m ¡°Þ°×°¿¤éãé F ÛØéËÍØéãé F>+
~¤Ë¤é د¤éÅà±ÅÛÅé°Ëؤ¬¤Õé v ×Û¯éدØé Fw °×éé¥ØÍÔéÍ¥é 8
=
?
D
v
8x
D M
?? M
8?= L
ID8 M
?=8 8x F44)
5L, kË×ߤÔYé KL*
¤Øé aj ¤éد¤é ͤ¥ª±°¤ËØéÍ¥é ~i °Ëéد¤é¤àÏË×°ÍËéÍ¥é 8= ~ :55& ¯¤Ë
844 ai 8x =i £Â #
¯Û×é @oa
8x
@o
Ϳޱˬéد¤é±Ë¤ÒÛ¿±Øãé
=
846 Ax 8
*
844 H
=844# y :
4 y8 ¬°Þ¤×éد¤é×Í¿ÛØ°ÍËé y 3
8MM3?& Øé°ÅÏ¿°¤×éدØ
~¤Ë¤éد¤é Ë×ߤÔé°×é KL$é 5O* kË×ߤÕYé ©. ÍؤéدØ
¯Û×é
¯Û×éد¤éË×ߤÔé±×é ©ı
OOÂ
=844# H y8
'+ '+ !+
=44K °×
5R* kË×ߤÔZé ÍؤéØ®Øé ı ı,ðı }ı 5&Hı7&Hı8&H !ıı9&H ı,ı , 5 ııııı,ıııı ıııı,ıııı ıııı,ııı ıııııı ı ı ıı ,ıı9ı9ı9ıé* °Ë¤é , ı~ı BÐ , ı Iı ×Íé ,ı ı+!h °Å°ÀÔÀãé ߤ鬤Øé ı+!#ı+!ıË¡é+,!ı ®Û×é ߤé®Þ¤é !ıı !ıı!ıı¬§ıı,&ı { ıı ı , rr 9ı ×Íé 5.'H ı !ıı!ıı!ıı 9&H ı ı
ıgé . =4$ kË×ߤÕ_é 54, ¤é®Þ¤
"6H "6H RG^x " 6=
d )
"H 6=H
®Û×é 5#@@DH ı__Nı ~ı 9ı9ı9ıı Nı ı %+ <5* kË×ߤÕYé 5KKOL* ¤é Åãé ÏÕͤ¤¡é ãé Ûװˬé Ø®¤é ÏÕ°Ë°ÏÀ¤é Í¥é °ËÀÛ×°ÍËé Ë¡é ¤àÀÛ×°ÍËé ×é ¥ÍÀÀÍß×$é
¤Øé Ø®¤ ÛË°Þ¤Õ×Àé פØé Gı ¤é Ø®¤é פØé Í¥é O ¡°¬°Øé °Ëؤ¬¤Õ×é ÍÅÏհװˬé Ø®¤é Oé ¡°¬°Ø×é ¥ÕÍÅé 5é ØÍé O*é
¤Ø Zĝı  Xı  :ı ¤é Ø®¤é ÏÕÍϤÔØãé Ø®Øé Xııı °ÅŤ¡°Ø¤Àãé ¥ÍÀÀÍß×é Xı °Ëé Ëé ¤À¤Å¤ËØé Í¥é G ~¤Ë¤é Bı ı Dı Xı ı ąÈ+ Xı ¥ÍÔé ı ·Â Xı  :^ı ®¤Ëé Ø®¤é Ë×ߤÕgé Bıg »´Bı »´ııı 9ı9ı9 ı @ı ı z :.ı ®¤é Ë×ߤÕé Ëé À×Íé ¤é Íذˤ¡é ¥ÕÍÅ ¤°Ø®¤Ôé Íˤé Í¥é Ø®¤éØßÍé¤àÏÕ¤××°ÍË×éËŤÀãé #+ íı ÍÔé ëı ı ,é ß®¤Ô¤é dı°×éØ®¤éËÛŤÔé Í¥ ¡¤Õˬ¤Å¤ËØ×é Í¥éØ®¤é°Ëؤ¬¤Ô×é¥ÔÍÅé5éØÍé +0 Ë¡é dı~ııĖz+ Ùıı $++ "+ı9ı9ı9ııızıdı* ==$ kË×ߤÕZé :ı.
nÆO Éı
E# )HE# *HE# +HE# H"H#H_2Â)H
E# A E# *E# +E# & "H"H#H
E# @2(H/B# )HE# +H #"H"H#H *E# ı0C# &E# )E# "H E# ı0C# 1(H )&#"H
,H#)H -H+ ) ,H+ +#)HH
®¤Ô¤¥ÍÕ¤é Ja ı 4
:ı =B* kË×ߤÔYé My ¤é ×®ÀÀé Ûפé Ø®¤é DCa LI_aJ\RH[MTRa ß®¤Ô¤é DCa ×ØË¡×é ¥ÍÔé ͬ¤Õé Í®é ØÍé ¡¤Àé ß°Ø®é Ø®°× ÒÛ¤×Ø°ÍË$é
¤Øé L4#ı'Jı=ı ¡¤ËÍؤé Ø®¤é ËÛŤÔé Í¥é ϤÔÅÛØØ°ÍË×é Í¥é 4ı ÍÏ°¤×é Í¥é 3H 'ı ÍÏ°¤× Í¥é ı Ë¡é =ı ÍÏ°¤×é Í¥é ı ×Û®é Ø®Øé ËÍé ØßÍé ¡»¤ËØé À¤ØؤÔ×é Õ¤é °¡¤ËØ°À$é kÅÍˬé Ø®¤×¤ L4Jı'Jı=ı ×Û®é ϤÔÅÛØØ°ÍË×é À¤Øé L4ı' #ı=ı ¡¤ËÍؤé Ø®¤é ËÛŤÕé Í¥é Ø®Íפé ϤÔÅÛØØ°ÍË× ß®°®é¡ÍéËÍØ餬°Ëéß°Ø®é 3H ~¤Ë¤é ߤé®Þ¤
L4#ı'Jı=ı L4'J=ı
Lı4ı+ |ı'Jı=ııLıı'ı+ zı4Jı=ıı $ ıp
+ |ı4Jı'ı L' |4Jı=ııL= |4Jı'Eı
OSÂ
®Â
yÍÔéÙ®±×éÒܤ×Ù±ÍÈé ߤéÔ¤éÔ¤Òܱդ¡éÙÍé¤ÞÀÜÙ¤é ıııı
ı ııı ııı ııı ııı ııı ııı ı ııı ııı ııı ııı ııı ııı
ııııııı ı#ıı 8x °ııı 8x ı ıı ıÃııııııÄıı=a (ıı ı=a (ı ııııııı 8x ıı@ı 8x (ı ıııııı 8x ııı=a ı ıı 8x ı ıııııı=a ııııı 8x @ı ıııııı 6x ıııı=a /ı ııı=a ııı=a ı ııııııı 8x /ııı=a ı ııııııı 8x ııııııı ııı°ııı 8x ıı:ı=a ı ıı(ı 8x /ı ııııııı=a ııııııı ııııı 8x ııı 8x ı ıı ı 8x Sı
ııı
~¤È¤é
?:aı :a 8x YP
ı
ı=a ı
ı
oãé <é ß¤é ®Þ¤é ı 6x+ı Ü×Ù±ÙÜÙ¤é Ù®±×é±ÈÙÍé 5éÈ¡é×±ÅÏÀ±¥ã ?;aıP O?:- ı 8x Ú ~¤È¤é : ? =
=a
ı
ı
0'KÂ
®¤Ô¤¥ÍÔ¤é ı=a
7ı .?- ı=a
Cx
*
+
*
~¤È¤é
$ * 5:a ı
# * 5:a ı
®Ü×é ı 8x ı ı6x ıı=a /ı È¡é ıoB ıoBBĘı 8x /Čı
OUÂ
#
5:a
#
5:a
ÿ"lGi& - +
+
+
& 3 ÅFı ı &ĕ F F 3 &ı ı ı1ıı ı1ı ıı ı1ı ¹ı Ć&ıı& &ı0ı1ı <ı <ı1ıı F&ı& &ı ı 0ı1ı <ı <ı1ı<& 1ı0ı <ı1ı 0ı& 1ı 0ı 0ı3ı 1ı ı <ı<ı ı3ı3ı ı 0ıı0ı3ı 3ı ı ı ¬Â 0ıı0ı1ı ı1ıı3ı 0ı & 1ıµ3ı 0ıı0ı3ı 1ı &&ı ı 3ı 3ı ı ¬ &&ı1ı ı1ıµ1ı 0ı 1ı ı3ı 0ıı0ı1ı ı3ı¬ 0ı 0ı3ı 1ı ı  ć&ı ı aııp aı aı aı p aı p aı p ııı K
Ġ ıı ı p ê ı :¡ı
OWÂ
+
+ + +
966
$ !$ "$ $ !"#$ QÂ "#$
$
!"! $ !$ ! !! $ TÂ !(#'#.
("
#+". .
.
+. . '. #' #. . ,(". +".
G
. !(#'. ""#. (%
".
. . ( '"#. ". +.
5, y°Ë¡é¿¿éÏ°Õ×éÍ¥é ÏÍ×°Ø°Þ¤é °Ëؤ¬¤Õ×é Ûı¸ı ×Íé Ø®Øé ııc ı 6x + <, Ëé Ø®¤é Ûؤé Øհˬ¿¤é ` rel="nofollow">ı k °×é é ÏÍ°ËØé °Ëé د¤é °ËؤհÍÕéÍ¥é د¤é פ¬Å¤ËØé `ı Ë¡é $ °×é ÏÍ°ËØé ÍËé Ø®¤é ¤àؤË×°ÍËé Í¥é د¤é פ¬Å¤ËØé `ı ×Û®é Ø®Øé ¨Â$ `ı
¤Øé ı Ë¡é ı ¤é Ø®¤ ¥¤¤ØéÍ¥éد¤é ϤÕϤˡ°ÛÀÕ×é¥ÕÍÅé krÂ Ë¡é ©Â ÍËØÍéد¤é¿µË¤×é `ı Ë¡é ı Õ¤×ϤذޤÀã-é ÕÍÞ¤ Ø®Øéد¤éÍÕدͤËØÕ¤éÍ¥é ¨ `ı ¿°¤×éÍËéد¤é °ÛÅ°ÕÀ¤éÍ¥é ¨ . +
B*
¤Øé /x Bx ¤éÏÍ×°Ø°Þ¤é°Ëؤ¬¤Õ×é߰دé q
9
bx A Uı Ë¡é߰دé qı ¡¤Ï¤Ë¡°Ë¬éÍËé wı tÍË×°¡¤ÕéØ®¤ פÒۤˤé2ıARı2ÍıE E E #ıı2ı ß®¤Õ¤
2dı
¥ÍÕ·gé #ıE E E eııı gé 6éxxxı#ıq+ + 2!ı2 ı !fı ¥ÍÕé bı ı æĞAı2
߰دé m_x+ qı+ ıgé ı ÍÕé <é °,¤,é 2ĥıgé 2fcı2 ı fcı ß®¤Õ¤é ¸ıgé ı ĦfcfAı2 ÍÕé <$é ¯Û×é °¥é .+x 6x UDı د¤é פÒÛ¤Ë¤é °×é Rı#ı#ıeıU#ı #ıUı Ë¡é °¥é ıgé >ı Ø®¤é פÒÛ¤Ë¤é °× 6é#ıRı5ıUeı Rı:Rıeı #ıU#ı.Qı y°Ë¡éGıgé2Aı2 ı ırrr2ı °¥é °Qı ıgé :Cı °°a + ^ı +
Cé
¤Øé ı 3
2eı ı 3
0x.ı ®Íßé Ø®Ø
Uı 9
:2ıı ę2xı ı ı2ı Ěı2ıĬěı ıĭı ı+ I* tÍË×°¡¤Õé é ı ~
ı ®¤××é ÍÕ¡(é
¤Øé k/ ¤é Ø®¤é ×Å¿¿¤×ØéËÛŤÕé Í¥é Õ¤Øˬ¿¤×é Ø®Ø Ëé¤é¡ÕßËéÍËéد¤é¯¤××éÍÕ¡é×ÍéدØéØ®¤é×°¡¤×éÍ¥é¤Þ¤Õã餿¿éÍ¥éØ®¤éÍÕ¡é°×éÍËذˤ¡ °Ëé Ø®¤é ×°¡¤×é Í¥éÍˤé Í¥é د¤é Õ¤ØˬÀ¤×*é y°Ë¡é Ø®¤é Þ¿Û¤é Í¥é ls! yÍÕé ¤àÅÏÀ¤ é ¥ÍÕé د¤é ı ~
¯¤××éÍÕ¡ é د¤éÞ¿Û¤éÍ¥é  °×éB$
OYÂ
+ + +
966
$ !$ "$ $ "! $
¥éWa°×ééÏԱŤé¥ØÍÔéÍ¥é +0 0 °Øé°×é¿×Íéé¥ØÍÔéÍ¥é +%00 $ ~ÍߤޤÔé ÈÍȤéÍ¥éØ®¤éÈÛŤÔ× 00 0+0 °×é é ¥ØÍÔé Í¥é +%00 $ Íé ߤé ÅÛ×Øé ®Þ¤é Wa 8x +00 $ °$¤$é +0  0 ÅÛ×Øé ¤é ÏԱŤ$ yÍÔé +0  0 6x ; éı/Cı ß¤é ®Þ¤é×Í¿ÛرÍÈ×é + 0#0 8x 0 0 ; é 0 È¡é ı; $é
¤ØéÛ×é×®Íßé Ø®Ø Ø®¤Ô¤é Ô¤é ÈÍé ÍØ®¤Ôé ×Í¿ÛØ°ÍÈ×$é È¡¤¤¡é ×ÛÏÏÍפé Ø®Øé +0  0 °×é é Ï԰Ťé ÈÛŤÔé rel="nofollow">?é :ı ®¤È + 6x ;Æé Æé ;
;$é Íßé ;é 3
Æé 3
+0 È¡éØ®Û×é +0 8x ;ÆÉé¡°Þ±¡¤×é +%0 ®Û×é +0 ¡±Þ±¡¤× +0 $ 0
+
+$0$#+$&0 Â
+
+
+
##0 0 +0Â +# ;é
$
®Û×é +0 4H #+0 È¡é °ÈéÏÔرۿÔé #0 CD +0 Øé¥Í¿ÀÍß×éØ®Ø ,0 +%0 8x +0 $ $ 0 ;
+$0 EFÂ + >?é +%
éÍÈØÔ¡±Ø±ÍÈ%é
<$
¤Øéé¤éØ®¤éÏͱÈØéÍ¥é°ÈؤÔפذÍÈéÍ¥éué È¡é é ı Øé°×é¤×ãéØÍéפ¤éØ®ØéØ®¤é±ÔÀ¤é߱خ ¡°Å¤Ø¤Ôépé°×éØ®¤é°ÔÛűԿ¤éÍ¥é W Üı k×é lé±×éÏÔÀ¿¤¿éØÍ é éÈ¡ér°×éÏÔ¿À¤¿ ØÍé ß¤é ®Þ¤é lré 8x é È¡é lré6x $é ±È¤é lré 6x é ߤ Ø®Û×é®Þ¤é w lé ré°×éÍȬÔÛ¤ÈØéØÍé wé é $
®¤Ô¤¥ÍÔ¤é Ø®¤é ¡±×ØÈ¤é ¥ÔÍÅé é ÍÈØÍé lré ¤ÒÛ¿×é ØÍé Ø®¤é ¡°×ØÈ¤é ¥ÔÍÅé éÍÈØÍé lr$é oÛØé È¡ééÔ¤éÍÈéØ®¤é׍é×±¡¤éß°Ø®éÔ¤×ϤØéØÍéØ®¤é¿±È¤é lré ±Øé¥Í¿ÀÍß×éØ®Øéé é°×éÏÔÀ¿¤¿é ØÍé lr$é ®¤Ô¤¥ÍÔ¤é é é ±×é ϤÔϤȡ°ÛÀÔé ØÍé péÈ¡é éÀ°¤×é ÍÈé Ø®¤é ±ÔÀ¤é ß±Ø®é ¡±Å¤Ø¤Ôé pé°ÔÛÅ×Ô±±È¬éÍÛØé W ı
B$ ¤é ×®¿Àé ×ÍÀÞ¤é Ø®¤é ÏÔÍÀ¤Åé ¥ÍÔé ¬¤È¤Ô¿é +0 ȱرÀÀãé À¤Øé ! 0! 00 0!,0 ¤é Ø®¤é FGZM]Ia פæ ÒۤȤ$é kÈéTWIXFZMTRaÔ¤ÅÍÞ¤×éØ®¤é¦±Ô×ØéخԤ¤éؤÔÅ×éÈ¡éÏϤȡéØ®¤±Ôé×ÜÅé×éØ®¤éÀ×ØéؤÔŠͥéØ®¤éפÒۤȤ$é ®Û×é¥Ø¤ÔéÍȤéÍϤÔØ°ÍÈéÍÈéØ®¤é±È°Ø±ÀéפÒۤȤé Ø®¤éØ°Þ¤éפÒۤȤé°× ! 0!. 00 0!,'/0 Û®éÍϤÔرÍÈ×éß°À¿é×ØÍÏéß®¤ÈéØ®¤éرޤéפÒۤȤé±×éͥ鿤ȬخéØéÅÍ×Ø <$ Èé Ø®±×é פé Ø®¤é À×Øé ؤÔÅé Í¥é Ø®¤é Ø°Þ¤é פÒÛ¤È¤é ±×é !*0 ®¤é¥Í¿ÀÍ߱Ȭé ÍפÔÞØ°ÍÈ× Ô¤éÍÞ°ÍÛ×$é
OZÂ
^ m§Ø¤Õ餯éÎϤÕØ°ÎÈ é د¤é¿¤È¬Ø¯éΧéد¤éØ°Þ¤éפÒۤȤ顤Ԥפ×éãé <éÈ¡éد¤é×ÛÅ Î§é °Ø×éؤÕÅ×é°×é ÏդפÕÞ¤¡& <- §é د¤é ¿¤È¬Ø¯é Χé Èé Ø°Þ¤é פÓÛ¤È¤é °×é ?n È¡é د¤é ¿×Øé ؤÕÅé °×é Nk
د¤Èé §Ø¤Õé ¡ ÎϤÕØ°ÎÈ×é د¤é ¿¤È¬Ø¯éΧéد¤éØ°Þ¤éפÒۤȤé°×é " È¡éد¤é¦²Õ×ØéؤÕÅé °×é Pl`
ÎéÎÅÏÛؤé XÂ ß¤é ¯Þ¤éØßÎéפ×W s×¤é ² 'é dx Ρ¢(é
¤Øé  Ȣé (, ¤é°Èؤ¬¤Õ×é×Û¯éدØé [x 8x ?k B=z 3
?kt °*¤* é , { 3
?k_ Îؤé دØé [»±Â 3
[%x ¯¤é ¦°Õ×Øé w±Â ؤÕÅ× é $B
P2/
°×é ¿¿¤¡é د¤é !" " È¡é د¤é ×ÛŠΧé°Ø×éؤÕÅ×é°×é  8 B, B B?z 8x ?|?{ B ,3 B Îßé ÏÏ¿ãé د¤éÎÞ¤éÎפÕÞØ°ÎÈ×( ¯¤é ×ÛÅé Χé د¤é ؤÕÅ×é Χé Èãé Ø°Þ¤é פÒÛ¤È¤é °×é 6 7a B , B B Wx 8x \ [x B 3( m§Ø¤Õé kx ÎϤÔØ°ÎÈ× é د¤é ¿¤È¬Ø¯é Χé د¤é Ø°Þ¤é פÒÛ¤È¤é °×é ?k È¡é °Ø×é ¦²Õ×Øé ؤÕÅé °×é $A9 0@ ¯°×éØ°Þ¤éפÒۤȤé°×é¿¿¤¡éد¤é ¶¹½ª«´¦Â " m§Ø¤Õé % &, ÎϤÕØ°ÎÈ× é د¤é¿¤È¬Ø¯éΧéد¤ Ø°Þ¤éפÒۤȤé°×é ".0#B ¯°×éØ°Þ¤é פÓÛ¤È¤é °×é¿Á¤¢éد¤é ¸³Â " ¤Ï¤Øéد°×éÛÈØ°¿ د¤é ¿¤È¬Ø¯éΧéد¤é Ø°Þ¤éפÒۤȤé°×é . ¯°×é °×éÈÎßé د¤é ¼é B+', ¿Î½)é ¯Û×é ß¤é ¯Þ¤é ¿Î½é ߯Îפé ×ÛÅé °×é uvÂ È¡é ¡Â B 7a ¿Î½×é ߯Îפé ×ÛÅé °×é ¼Â ~¤È¤é Gı /  " B , °È¤é d  8x J=Q! Yx 3x , 8x fg B=rhUCM ! ߤé¯Þ¤éGı 8x <2HRD?FRM& s×¤é ²°& Xx ¤Þ¤È)é
¤Øé ¢Â È¡é (, ¤é ³Èؤ¬¤Õ×é ×Û¯é دØé cx 8x =?l B (, 3
=?m; ²*¤$ +
+ + +
+
+ + +
+
(Â }Â jx 3
%, ÎؤéدØé ]J²Â 3
Z$x ¯¤é¦²Õ×Øé \I±Â ؤÕÅ× é P'
O1n
°×é¿¿¤¡éد¤é !" " È¡é د¤é ×ÛÅé Î§é °Ø×éؤÕÅ×é °×é t  m×é °Êé ×¤é ° é §Ø¤Õé +é ÎϤÕØ°ÎÈ×é د¤é ¿¤È¬Ø¯éΧé د¤é Ø°Þ¤ פÓÛ¤È¤é °×é =VCo È¡é °Ø×é ¦²Õ×Øé ؤÕÅé °×é cW}s ¯²×é Ø°Þ¤é פÒÛ¤È¤é °×é ¿¿¤¡é د¤é " " m§Ø¤Õé =%, ÎϤÕØ°ÎÈ× é د¤é ¤ȬدéΧéد¤éØ°Þ¤éפÓۤȤé°×é =[Cm%^ - ¯°×éØ°Þ¤ פÒۤȤé°×é¿¿¤¡éد¤é ! " " ¤Ï¤Øéد°×éÛÈØ°¿éد¤é¿¤È¬Ø¯éΧéد¤éØ°Þ¤éפÒۤȤé±×é<(é ¯°×é²×éÈÎßéد¤é Na B)#, ¿Î½(é ¯Û×éߤé¯Þ¤éé¿Î½é߯Îפé×ÛÅé°×é jq ȡé  B, ¿Î½× ߯Îפé ×ÛÅé °×é  ~¤È¤é Gı x¥Â B / B , °È¤é C g 8x , *$, ¯Û×é x 8x , È¢é  BÈ¡é×Îé ~ " ,
+
+
+
+
E0 z°Õ×ØéÈÎؤé دØéãé k!| '3:B tx µ1 +
 ;+1Bc:,2B%B(4<B&B '5=BP
7
= =,1B%B'6=B"
+
¯Û×é yoÂ
B r*] ux r{D S
Lhq S
'6=B Â p + V +
P
8x +
, B,
!, ,
7
#,x :=?x x
x
x
F\]D
)
Ex s\f S
Xet
Ox
B !, , , B
+ ,x >TUKx x
Mis-x gx
!"x
t\f*
S
Wfy [
x
_
B ze-
x x x v x ydTXjyT
[Wiu,
T
+ , , B @QQPx
+
&
<x
#H
x *+
7
, 8
= '6> c B >,2$
¯¤é¿×Øé°È¤ÒÛ¿°Øãé§Î¿¿Îß×é¤Ûפ鰧é ;+1?B8x ?3F د¤Èé'6>? =,2B1lx )6>B$B B'7=B?B>+2B$B
S)Â
+
D 4 , È¡é -Â 7Â =+2$ ?B 8x
I* ¯¤é È×ߤÕé °×é j $ m¿Áé د¤é ¯ÍÕ°åÍÈØ¿é ×°£¤×é Èé ¤é ÍÞ¤Õ¤£é äé 544Eé Ï°¤¤×é ͧ + }
$ Õ¤ØȬ¿¤×é ¤à¤ÏØé د¤é ÍÝÈ£Õäé ͧé د¤é ¯¤××é ÍÕ£é ߯°¯é Èé ¤é ÍÞ¤Õ¤£é ä د¤é ÍÝÈ£Õäé Õ¤ØȬ¿¤&é ¯¤é դŰȰȬé Þ¤ÕØ°¿é ×°£¤×é Èé ¤é ÍÞ¤Õ¤£é äé 544Eé Ï°¤¤×é ͧ $ }
$ Õ¤ØȬ¿¤×*é ¯Ý×é ip ¾Â
+
Íßé×ÝÏÏÍפéدØéد¤é¯¤××é ÍÕ£é¯×é ¤¤ÈéÍÞ¤Õ¤£é äé h դØȬ¿¤×é°Èéد¤é£¤×°Õ¤£éßä$
¤Øé $ ͧéد¤éÕ¤ØȬ¿¤×é¯Þ¤éد¤°Õé ØÍÏéÈ£é ÍØØÍÅéÍÈé د¤é ØÍÏé È£é ÍØØÍÅé ͧé Ø®¤é ÍÕ£ $ ͧéد¤é Õ¤ØȬ¿¤×é ¯Þ¤éد¤°ÕéØÍÏéÍÈéد¤éØÍÏéͧéد¤é ÍÕ£é V
ͧéد¤éÕ¤ØȬ¿¤×é¯Þ¤éد¤°Õ ÍØØÍÅé ÍÈé د¤é ÍØØÍÅé ͧé د¤é ÍÕ£é È£é $ ͧé د¤é Õ¤ØȬ¿¤×é ¯Þ¤é Ȥ°Ø¯¤Õé د¤°Õé ØÍÏé ÈÍÕ ÍØØÍÅéÍÈé د¤éØÍÏéÍÕé ÍØØÍÅé ͧéد¤é ÍÕ£( ´È¤éد¤Õ¤éÕ¤é <44Lé°ÈؤÕÈ¿é¯ÍÕµåÍÈؿ鿰Ȥ×é ߤé¯Þ¤é ııN x $ $ < $ °È¤éد¤Õ¤éÕ¤ =44RéÞ¤ÕØ°¿é¿°È¤×é°ÈؤÕפذȬéد¤éØÍÏéͧéد¤éÍÕ£ é ߤé¯Þ¤é$$ A $ hm
$$ > $ °Å°¿Õ¿äé Nx$ A $ ¯Ý×é !$ $ $$Jx ? $ ~¤È¤é $$$$V
$ = $ +* B j $$Nx $  °¤Â @Â
+
P,Â
8^oSBws|O2BXOkB^IBf 8sI^O h~¢¥¶b~ª~ª~¶e¯¢~¶hbe¶ D33W !" "
%?2%).%%Ã
u²®z»Ã.?Ãiz»Ã.%%;Ã
" (-1,4 4 FL± +/-.")(-
(.,4 2)/,4 (-1,-4)(4.!4 (-1,4 -!.4 *,)0" 4
),4 .!4 '/&."*&4 !)"4 +/-.")(-4 (.,4 )(&24 .!4 &..,-4 4
VÃ i£ k£ ),4 4 ),,-*)("(
4 .)4 .!
),,.4 (-1,-4 "(4.!4 (-1,4 -!. 4
),4 .!4 ).!,4 -!),.4+/-.")(-4 1,".4 2)/,4 (-1,-4"(4(-1,4 -!.4 (4 -!4 .!4 **,)*,".4 /3 &-4&)142)/,4 (-1,- 4
)4-.*-4,4 (4 .)4$/-.#4 2)/,4 (-1,- 4
!4+/-.")(4,,"-4 6± ',%
)4&/&.),-4,4&&)14
" " "" " " "" " " " "
7±
)" S¡¥¢Ã¯Ã¥²©Ã®¯z¯¡¢¯®Ã¥¢Ã¢¯©®Ã|¥¶ cà 4 4 4 ¯¢Ã 4 4 4
4 kà %îÃz¶z»®Ã¯©²L
4 4 %îÃz¶z»®Ã¯©²LÃ
cà 4 4 4 ¯¢Ã4 4 4
¥¶Ã¡z¢»Ã¥Ã¯¡Ãz©Ã¥©©¯Rà SÃ%Là WÃ)LÃ
j£.LÃ
[Ã2LÃ ]ÃG*£
. xåï奶¢Ã¢²¡|©®Ã®Ã¥Ã¥©Ãz¢»Ã¢¯©Ãµz²®Ã¥ÃR Sà .%%;à £3Là WÃ.%%;à £;Là ZÃ.%%;à £. Là [Ã.%%;à £.%%;Là ]Ã.%%; 2 c¢ÃzÃz 2%%Ãr¥¢z©»Ã2¢¯®Ã®¯²»Ã¯©Ãa¥©z§»ÃW¥¥»Ã¥©Ã|¥¯Ãa¥©z§» z¢ÃW¥¥»Ã cà >%à ®¯²»Ã a¥©z§»Ã z¢Ã 8%à ®¯²»ÃW¥¥»Ã ¥¶Ã¡z¢»Ã®¯²¢¯®Ã ®¯²» |¥¯Ãa¥©z§»Ãz¢Ã|¥¥»R SÃ2%Là WÃ:%Là ZÃ>%Là [Ã?%Là ]Ã)9%à 5 S¢Ã²¢|z®Ã®¸®Ãîⲡ|©Ã)ï¥Ã:à uÃîﬥ¶¢Ã¯¶Ãz¢Ã¯Ã¯¶¥Ã®¥©® zà xåï奶¢Ãµ¢¯®Ãz®Ã¯Ã®¯Ã§©¥|z|¯»Ã¥Ã¥²©©¢R SÃu說zîÃz穡ⲡ|©LÁà WÃu說zîÃzᲯ§Ã¥Ã5Ià ZÃu說zîÃz穯z©Là [Ãu說zîÃ;Là ]Ãu說zîÃzÃz¯¥©Ã¥Ã).à 8 uà z©|¥z©Ã|¥¶Ã z¢Ã |à ²¯Ã¥²¯Ãz¢Ã¥Ã¯¥Ã ¡zà zà ²|à xÃzà ¶Ã¯¢Ã| ¥§§¥®¯Ã¯Ãzáz©ÃTR
\Ã
i£ TÃ
VÃ
SÃV JÃ WÃi[£ ZÃ\IÃ [Ã^IÃ ]Ã
.Ã
^Ã
:" b¥¶Ã¡z¢»Ã¯©z¢®Ãz¢Ã»¥²Ã¢Ã¢Ã¯Ã¥¥¶¢Ã²©R 4
g£
k£
£
g£ n£ f [£ h 8[£ j B[£
[ 5IÃ ] 8
B£
W$
r²§§¥® )à l±
W±
8£ B£ = £ W± W± £ £ £
£ xz¯Ã®Ã¯Ãµz²Ã¥Ã c£
l±
f£)Jà h£8Z£ j£BZ£ )%à !Ãr²§§¥®Ã =
[Ã5IÃ ]Ã8
6B£ £)à W±/$£ xz¯Ã ®Ã¯Ãz®¯Ã¯Ã¥ÃF£ £ Fc
f£)KÃ h£BZ£ j£8IÃ
[ÃOZ£ ]ÃW$£
))" c¢Ãzï©z¢ÃTVYà ¯Ã®Ã µ¢Ã¯z¯ÃTVà 8T )áà VYà W± 8//O£¡Ãz¢ÃTYà W±x± ¡Ã ¶©Ã x±®Ãz¢ ¢¯©Ã [¯©¡¢Ã¯Ãµz²Ã¥Ã x!
£)8£
). _¢Ã¯Ãµz²Ã¢Ã¯Ã®¡§®¯Ã¥©¡Ã¥ )2 _¢Ã¯Ãµz²Ã¥
68£2
8//O=£ £8//V=£ 6WWB=£6WW8= EU
5à )5 _¢Ã¯Ã©z¯®¯Ã¢¯©Ã4®²Ã¯z¯
8//O/£ƇB)
4
)8 r²§§¥®Ã¯z¯Ã£z¢ºÃz©Ã¢¥¢¼©¥Ã©zⳡ|©®Ã®²Ã¯z¯ k±
l± W±=
z¢Ã A¶ W±?º
B£
_¢Ã¯Ãµz²Ã¥Ã £ º £ à ): ]µz²z¯Ã¯Ã®²¡
8//O£ 8//T£ 8//O£ %£ ! £ )à à 8£ 8£Ã B£ 8//K£Ã 8//O£
6O$
_¢Ã¯Ã®²¡Ã¥Ã¯Ã¯®Ã¥Ã¯Ã§©¥²¯ ) à ))à à 8//O£
= 22U£,
QT
5Ã
)> u
Ãz©z¡Ã®¥·®Ã¯·¥Ã
¢¯zz©
®Ã dgik£z¢ÃmopsÃ¥µ
©z§§¢Ã
zÃ¥¯
©Ã¢Ã®² z÷z»Ã¯z¯Ã¯
©Ã
®Ãz©
çz©z
Ãz¢Ãzé
åéz²®Ã . à j¶ ¡Ã¥µ
©
÷¯¢Ã¯
®
®¨²z©
®Ã _¢Ã¯
Ã
¢¯Ã¥Ã¯
z©
à dgik$
6\) x
¢Ã.%%<Ã|z©®Ã¥Ã®¥z§Ãz©
çz
⯥ÃcƇ|¥¸
®Ã¥Ã
¨²zî½
÷
©
ÃcƇ®Ãz¢Ã¢¯
©Ã®¯©¯» |
¯·
¢Ã.%%Ãz¢Ã2%%ï
©
Ãz©
Ã
¸¯©zÃ8Ã|z©®Ã©
¡z¢¢Ã _¢Ã cƇ ĚĐ r²§§¥®
ïz¯Ã £»Kà W±.%%:Ã
£ £»Kà W±.%%+Ãz¢Ã££»Là 7T .%%>Ãz¢Ã
£ W±2à _¢Ã¯
õz²
Ã¥
£
£ £ )Ã )Ã &Ã $£
£ £
£ £
£ £ .) u
Ãz©z¡Ã|
¥·Ã®¥·®Ãzï©z¢
à dgi£¢Ã·Ã qd£ W± :%¾ÃVmÃz¢Ãgm£¯©®
¯Ã qdgiZ£z¢ YmÃz¢Ã im£°©®
¯Ã qdig$£ f
¯Ã¯
Ãz¢
à qgxm£|
à k,± _¢Ã »&Ã
sÃ
¤Ã
8Ã
88*
r²§§¥®Ã¯z¯Ã£ £ 7T
8B*
b¥¶Ã¡z¢»Ã¥©©Ã§z©®Ã ¥Ã ¢¯©®Ã ££Ã¶©Ã /£ ]£ 8//V9£ £9£^£8//O9£ ££-Q
8G*
cã £ %¶ ££ 6T W£z¢Ã9£ ££ ££W±8O £¢Ã¯Ãµz²Ã¥Ã
8I'
S§§¢¢Ã¯©Ã¯®Ãz¯Ã¯Ã¢Ã¥Ã8//O £¥¢Ã¥|¯z¢®Ã¡Ã ¢¯©ÃÑƇ¥Ã®µ¢Ã¯®Ã c¢Ã¥©© ¿¯¥Ã¯Ã ÑƇ¯¥Ã|ïᢡzⲡ|©Ã¶Ã®Ãµ®|Ã|»ÃB £I£z¢Ã O£®¡²¯z¢¥²®»Ã¶z¯Ãz© ¯Ã¯©Ã¯®Ã¯z¯Ã¥¢Ã¶¥²Ãz§§¢Q
8K*
_¢Ã¯Ãz©®¯Ã¢¯©Ã£Ã®²Ã¯z¯Ã L19 5£ `à 8//O911S*
8O*
_¢Ã¯Ãµz²Ã¥
¶¢Ã£ W±
3*£_¢Ã¯Ãµz²Ã¥ÃH£ A£ A£ B£ £B9£ £* £ ]£ £Ã ]£ 8//V£®z¯®»Ã¯Ã ¨²z¯¥¢ %¶ £,
H£ KA£ 89£ £3V£ £8B£ 9 V£ £3I£
3W V£u1
8V*
_¢Ã¯Ãµz²Ã¥Ã £®²Ã¯z¯Ã¯Ã¯¶¥Ã¨²z¯¥¢®Ã9£ £3£ W±/£z¢Ã9 ^£/£zµÃ¥¢ ¥¡¡¥¢Ã©z饥¯
8W,
k⯩®Ã®¯z©¯¢Ã©¥¡Ã 3£z©Ã©¥²§Ãz®Ã¥¥¶®Aà 3 £B £I £O £W £3 3 £3B £3I £3O £3W £# * * ¶©Ã ¯Ã £¯Ã ©¥²§Ã ¥¢®®¯®Ã ¥Ã £Ã ¥ §¶ ¢¯©®#à b¥¶Ã ¡z¢»Ã ¥Ã ¢¯©®Ã z©Ã ¢Ã ¯Ã ®z¡ ©¥²§Ã¶Ã 8//O£|¥¢®Ã¯¥Q
B/*
c¢Ã tgi£rgdi£ 6T G{v*£k£®Ãz祢¯Ã¥¢Ã gi£®²Ã¯z¯Ã dk£®Ã§©§¢²z©Ã¯¥Ã gi$£ cÃgk£ 7T B ¡Ãz¢Ãki£ W± 8£¡Ãz¢Ã¯Ãz©zåïà tgi£®Ã£¡9'£`¢Ã¯Ãµz²Ã¥Ã* tÃ
pÃ
¥Ã
K£
2) c¢Ã hVYîÃ|¥¶ÃTVÃNà TYÃNà µ{²Ã¥ÃV\ÃÀ\Y
¶{¢Ã\ ®Ã{à §¥¢¯Ã¥¢ÃVYî²Ã¯{¯ÃT\Nà )à `¢Ã¯
TÃ
\Ã
VÃ
£
2. `¢Ã¯Ã{®¯Ã¯Ã¥Ã 89:-.N£ £) 22 `± ¯Ã¥¥¶¢Ã{©{¡Ã TVY\îÃ{{©Ã {¢Ã^îï⯩åï{©ÃTVY\à mî {à §¥¢¯Ã¥¢Ã{ö¯Ã{¡¯©ÃTVà oà ®Ã{祢¯Ã¥¢Ã{î¡©Ã ¶¯Ã{¡¯©ÃT\ i¥©¥µ©Ã o Tà {¢Ã m{©Ã ¥¢{©Ã¯{¯Ã®Ã ¯»Ã {©Ã ¥¢Ã¯Ã ®{¡Ã ¢Ã r²§§¥®Ã oTÃNà )5 ¡ÃTmà Nà 5:áÃ{¢ÃT^ÃNà »Ã ¡Ã `¢Ã¯Ãµ{²Ã¥Ã º%Ã
ST
|Ã
¥Ã
zÃ
25 `¢Ã¯Ã®¡{®¯Ã§¥®¯µÃ¢¯©Ã£Ã®²Ã¯{¯Ã££Ã £) £Ã £8£®Ãµ®|Ã|»Ã.5; 28 `¢Ã¯Ã{©®¯Ã¢¯©ÃcƇ®²Ã¯{¯Ã|¥¯ÃcƇ £5?:Ã{¢ÃcƇ £..5Ã{©Ã§©¯Ã®¨²{©®
;Ã
8^oSBws|O2BXOkB^IBf 8sI^O h~¢¥¶b~ª~ª~¶e¯¢~¶hbe¶ D33W !" "! "
) S¢®Aà W k¢»Ã ±
à z®±Ã ®±z±
¡
¢±Ã ®Ã ¥©©
±DÃ y-Ã ^Ã }
0à ¡§
®Ã 0à kà %Ã
¢
Ãyà ^à à _¥©Ã ¥±
© !î±z±
¡
¢±®Ã¥²¢±
©
¸z¡§
®Ãz¢Ã|
ñz
Ãz®Ãyà W±.Ã}à W±)IÃyà W±%Ãz¢Ãyà PÃ%é
®§
±µ
» . S¢®BÃ j W
z²®
Ã.0îÃz¶z»®Ã
µ
¢Ã±²®Ã.%%;à Ã.0îÃz¶z»®Ã¥Ã _¥©Ã¥±
©Ã®±z±
¡
¢±®Ã
±
©Ãà W±) ¥©ÃÃOÃ%õ
®ÃzÃ¥²¢±
©
¸z¡§
2 S¢®AÃ [ w®
Ãc¢²®¥¢Ãz¢Ã]¸²®¥¢Ãn©¢§
5 S¢®AÃ S u
ⲡ|
©Ã¥Ã¥²©©
¢
®Ã¥©Ã
zÃ
µ
¢±Ã®Aà )8ÃW £;Ã:Ãz¢Ã).é
®§
±µ
» 8 S¢®Aà [ e²®±Ã¡z¢
: S¢®Aà [ e²®±Ã ¥²¢±Aà fz|
à ±
Ã
¢±
©Ã +"± u
©
à z©
à :à ±©z¢
®Ã
à tnw[£ 2Ã
à twg\£ :Ã
tgk£z¢Ã)
à tgi%£ u¥±zAà ): ; S¢®Aà W g
±yOà ))¹ z¢ }à W±)2¹Ã x
zµ
y3 }3à W± y }3à r¡§»Aà 2y}y }à W±%à q
§z¢ yÃ}Ã|zâñ
©¡®Ã¥Ã¹Ã¶
Ã¥²¢Ã±
ñ©
饥±®Ãz©
Ã))Ã).Ãz¢Ã)2à u²®Ã¹,ÃƇ¹-à ù4à PÃ2: > S¢®Cà W r¢
ñ
Ãz©
zåî
±¥©Ãdkn£z¢Ã tgi£z©
Ã
¨²zö
Ãzµ
)Ã l± = Ã 0
DU
)Ã
:¶ no¶ < :¶ u
é
®²±Ã¥¥¶®Ã
>Ã
W$£ S¢®Aà Wà B£ 3£ 8£ 3£ _£ z¢Ã _£ £ £ £ £
¯®Ã²®Ã
£ £ £ £ 8£z¢Ã £3£_£B$£ 7T
k±
k±
k±
u²®Ã £_£/£z¢Mà _£>$£ j¦¯Ã¶Ã¢¯Ã²®Ã¯Ãz®¯Ã¨²z¯»Ã|²¯Ã £_£Z£_£/£z¢Ã£ 8T 8£ HT ®z¯®»Ãz梯¦¢®Ã
3/$£ S¢®AÃ [Ã &£
3£
W»Ãz®®²¡§¯l¢Ã £
k±
3£ _£3B$£ u²®Ã 9 £ A±
7T
k±
¶¦®Ãz®¯Ã¯Ã®Ã O$£
F 3B 8£ 3KO$£ r¡z©»Ã £ £ :T
:T
8T
3£
_£3KO 8 ;T
3 3$£ S¢®Aà 8//O$£ w®Ã£ ^à 8//O£ £3£z¢Ã 8//O£^à £ £3$£ 38$£ S¢®Aà K'£ w®Ã 8 3 £¿Ã 38£34¶_£¶!¶¾Ã B9 3B'£ S¢®Aà B////$£ w®¢Ã9
9£_ £
£8//V9£Q± 3WWB9£ G0/3£¹Ã 3I£z¢Ã 8//O9 3WW89£_£BWWW£Ã 4I$ 5T
u鮲¯Ã¦¦¶®Ã
3G$£ S¢®Aà 8//8&£ W»Ã¦¡§¯¦¢Ã¦Ã®¨²z©Ã 8//O9£ 8//O/£ £B 3 £_£8//O£
I9£ £K$£u鮲¯Ã¦¦¶®Ã
4I$£ S¢®Aà 883G$£ ]¡¢z¯¢Ã £ ¶Ã¯Ã B9£ _£V'£ u¢Ã £ _£8O£®¢Ã £ X± ¦¦¶®Ã
/$£ u²®Ã £ _£8 3VO$£ uà ©®²¯
3K$£ S¢®Eà 8//K$£ 3£ 3£ 3£ ¶Ã¯Ã w®¢Ã _ £ £ £ 3£ 3£ 3£ £ 3£ 3£ £ 8//K$£ 8//O $£ 3 £ £$ $ $ £ 8£ 8£ B£ >//K£ 8//O£
W£
8T
3P(
S¤®Aà 3V/KB$ r¤Ã¯ªÃ®Ã¤¥Ãz«ª»Ã¯®Ã®Ã®¡§»ÃX£¹Ã 8//O)£ kªÃ¥|®ªµÃ¯Ã®Ãz¯²z»
888£XXXXXXXXX£ XXX£PPP£ +U R
R
U
+U
K??QÃ X ÃPT
3V$
S¤®Aà 3$ $S®Ã¯Ãzªz¡Ã|¥¶Ã®¥¶®Ãx£`£ $¡¶z¤Ã}£ 4T wx
x}£ x£ £}£`£ 3T
8 "£ 8 "£ wx£`£ 8£ 8£ 8£
`£3)
Ã
±Ã
3X$
¡x$£ u²®
S¤®Fà 8VK) X»Ãz®®²¡§¯¥¤Ã 8//O I£`£c £¥ªÃ®¥¡Ã¤¯ªÃ$£ _z¯¥ª¼Ã 8//8£`£8£ 8//£^à cƇ^à B// £¯Ã¥¤»Ã§¥®®|¯»Ã®ÃcƇ`£8£ 3 3 £ 3B£`£8VK£z¤Ã £a£P$ PU
OU
8/$
PU
P£
QU
S¤®Cà 3* q¶ª¯Ã¯Ã¸§ª®®¥¤Ãz®
9£ £
9£ £9£
£ £
£ £
£ KÃ ££
£ £9£ £££ £9 `£
£ 8
£ r¤Ã
£`£ 3 £
£`£ 3 £z¤Ã `£8 £¯Ãª®²¯Ã¥¥¶®Ã 3/£
QU
Ã
NU
3B$£ r¤Ã
8 3)£ U¢®Bà I/$£ 8£ 8£ pgxi£`£3V/ pxgi£ £pxig£`£3V/ pdgi£ pdil`£3V/ D 38/£`£3//* k|®©µÃ¯{¯Ã xm£|®¯®Ã pgxi £¯Ã©®²¯Ã¥¥¶®Ã
88$
U¢®Bà 3$ xá{¢§²{¯Ã¯Ã¸§©®®¥¢Ã{¢Ã©§{ã £|»Ã 3£¶¢µ©Ã¢®®{©»B
F£ C£ A£ B= £ £B=£ ££ A£ B £ `£ ;£ ££ £ B `£ =£ 3$£ *U
3T
3T
8B$
U¢®Bà B$ r¢Ã 8//V9 8//O9£ 2T =£ = £¶Ã{µÃG/3I£3T £Ã I£ 33£OB£`£ $ u©©Ã¥²©Ã§¥®®|¯®Bà £ £ £ b£ G/3I £ £`£ 3Z££ £ £`£ V/B £ £`£ I\££ £ £ b BKI £ 3T 3 3 £{¢Ã £ £ £`£ OB £` II$£ W²¯Ã ¯Ã ¬®¯Ã ¥¢Ã £`£ 8//V £ £`£ 8//O£® ©²Ã ¥²¯Ã|»Ã{®®²¡§¯¥¢Ã¯Ã ©¡{¢¢Ã§{©®Ã {©Ã £`£ G/G ££`£BWW[££`£ 3VV ££`£ 3OO£ {¢
£`£KG £`£W' .5 U¢®BÃ B$
r¢Ã £ u鮲¯Ã¥¥¶®
8I$
£`£ =£=£`£= £¶Ã{µÃW £ 8O& 2T
U¢®Fà /OI$ d¯Ã®²®Ã¯¥Ã¡{ïõ®|Ã|»Ã 3/I£{¯©Ã{§§¢¢Ã U®Ã8//O///£`£3/I£¹Ã 3W3 3G£ £B/ £¯ {®¯Ã¢²¡|©Ã¯{¯Ã¶Ã¢Ã¯¥Ã{îà OI$
8K$
U¢®Aà 38+ xà ¢Ã ¯Ã¡{¸¡{à £®²Ã¯{¯Ã E?11Q£ ^à 8//O?11R$£ xà ¢Ã E£ ^à 38A£`£ 3O8V£^à 8//O£^à 8 3WO£`£3BA$£ u奶®
33£
8//O$£W»Ã{²{¯¥¢
.; S¤®Aà 8 k~®ªµÃ£ 2T ¥¤Ãµ®¥¤Ã
#9£ W±G ¶z¤Ã9 V£ £)2ÃW±)? V£ 2.Ãà V£¶Ã )2à W±%à w®Ã F KA 8;£ £3V£ £.2Ã
2T
V£ £)2C¿Ã .£ £)Ã £)%
C
v²®Ã¯Ã¥ª¤zø§ª®®¥¤Ã¨²z®Ã 1 V£ £)2Âà £8£ £)à £)%Ã
9£ V£ £)2Ã £.
)%Ã
W±
.Ã
W±8
MU
.> S¤®Aà . w®¤Ã £)à W± w± ¶ÃzµÃ £ 6T )Ã¥ªÃ £ W±)ÃW²¯Ã¶¤Ã £ W± )ï媤zè²z¯¥¤ z®Ã¤¥Ãªzꥥ¯®Ã v²®Ã £ W± )öÃzµÃ £ W±. .? S¤®Aà 58 .%%;îïÃ)Ã%%5¯Ã¥Ã¤²¡~ªÃ cÃ.%%;îäïꥲ§Ã£ £)ï¤Ã)à £.à £ )à £.à £* ã £)à v²®
MU
£ £)Ã
_Ã )%%7Ã
££_Ã )%%5Ã GHÃ
MU MU
£ £4¥ƇƇ.
.à .à xïà Ƈ W±55à r¥Ã.%%;îäꥲ§Ã58öÃz®Ã58å䯪® _Ã
2% S¤®Aà )8 Z¥¤®¯ª´¯Ãim£¶Ã®Ã§ª§¤²zªÃ¯¥Ã eg£¶ªÃ m£®Ã¥¤Ã eg*£ r¤Ã pgei£ 8T 5| DT µ¤ em£W±im'£ v²®Ã¯Ãª¯Ã tmo£®Ã¥¤ª´¤¯Ã¯¥Ãª¯Ã 7img£r¥Ã eo£W±ig£W±8 j¸¯Ãtkg£®Ã®¡zªÃ¯¥ÃMiko £¯²®
gk£ ok£ ek£ eo£ 2à ek£ J£ W± ¶Ã¡§®Ã W± W± . ek£ ek£ ik£ ¶Ã¯Ã ek9£ |ek£ K£W±%Ãt¥µ¤Ã ek£ K£¥¤»Ãz®Ã ek£kà %à vꮲ¯Ã¥¥¶®
8T
T T T
à ~à à à à à T
à T T
oÃ
).Ã
\!ƇƇ! ƇƇ $¹$@Ƈ aƇ@! !k@*ôƇ! Ƈk$*Ƈa$Ƈ$ Ƈ $Ÿ$Ƈ!! Ƈ$Ƈ!!`Ƈ ÎƇ à $ ċƇ !Ƈ pldk£$ Ƈ pkdi£@mQ*ÃƇ r Ƈ$ Ƈ £¿ W±$ "ńĿƇ 7T 4Ƈ! Ƈ
)T CT
9T
W±4&
mynyÃ
Q! ? *Ƈdk£ W±K' «4 ) ĥƇ<$ ®! @žQƇ$ƇQ@QƇaƇd£$ƇaƇQ @ƇŐ Ƈdg£W±di£ W± m¶$ƇaƇ@$ ƇijR Ƈdk£!Ƈk aƇQ@Qk@ QƇ$Ƈ s£$ Ƈu£$Ƈa! Ƈ
(T ' &T %T @AT >T ?T
à à à Ã
à 0123à à à à à à à à à à Ã
'()Ã*+,Ã
(U
T
à à KT JT
à à à à Ã
tF¶
C¶
s¶
à à à à à Ã
Ã
à à à à à à à Ã
?T >T
JT
tÃ
T
*T 'T
xÃIT T
IT
*+,-./Ã
uÃ
' (T
4567Á:Ã
Ã
r ƇaƇĸ @Q Ƈ ®a!@ ƇUa!@kƇ
gk£ki^£sk£uk£ W±Ƈ# 4¥Ƈ"¶ £4¥Ƈ W±<$ « ) `Ƈ O$ P@Ƈa$Ƈ!@Ƈ ¬Ã @±< £<=£$$*Ƈ Ƈ aƇ$Ƈ K$£ UaƇ@Ƈ!!
4«Ƈ
22 ) ĦƇ 25
ĺ! Ƈ yk£ $ Ƈ xg£ Ƈ Ƈ $*Ƈ xg£`£yd£`£ & 5
!Ƈ !Ƈ $ Ƈ @ Ƈ ¢kyd£ Ƈ Q! @ Ƈ !Ƈ @ Ƈ txg%£
)mmZ*Ƈł* $!@$åƇ !@kƇ !Ƈ txg
£9£`£5:-Ã £ & G 9 ²!ZƇ £`£25
tÃ
à à à à à à à à à à à Ã
ST
|Ã
¤Ã
xÃ
25 ) `Ƈ 2;
² QƇ .5;Ã`£ &2 & ?à ! Ƈ!Ƈ ¬Ã ¬Ã £ &ìà £8£Ƈ ZƇ*Ƈ &2à Q$ZZƇ Ƈ£$ Ƈ! Ƈ*Ƈ & ? ÃQ$ZZƇ Ƈ
$ ®Z$@Z*Ƈ
£ £ £± . LU
ľ Ƈ
£`£ & ?
à ³ Ƈ
¿`£ &Ã QƇ
£ £± .Ã £ Ƈ $k! Ƈ & ;Ã& >Ã& ?Ã.%Ã. & Ã ; Ƈ ! Ƈ
ZƇ*Ƈ &2Ã QƇƇ @*Ƈ
¿`£ .Ã \!Ƈ
£`£ 2>à $ Ƈ £Ƈ $k! Ƈ 2:Ã2;Ã2>Ã2?Ã5%à Q `£2?à I Ƈ ƇZ$ Ƈ¬Ã`£2; 28 ) `Ƈ 5.:8
Î Ƈu£Ã 5?:à 0T 9£$ Ƈu£ £./5à 1T
=£!@Ƈ!kƇm! Ƈ Ƈw±$ Ƈ
$£ Ƈ @
@£`£5@:à 7T ..5Ã`£ .;.Ã`£ .6à & ;à I Ƈ &;'yà £
$£ {Ƈ & ;'y y± Ƈ
£ BC± & ; Ã$ Ƈ £
£ £± & : km!ZÃƇ Ƈ & ;(y à £
% OU
³Ƈ $ƇřťƇm!Z Ƈ!@Ƈ £
£
Y£ &;Ã& :Ã25Ã>Ã:>Ã5Ã &2:Ã.Ã.;.Ã& Ã ²!Z $ Ƈ Q$@ Ƈ ! Ą @Ƈ!Z ! ƇƇ $Ƈ £
£ 5T . & Ã&2Ã2:Ã2.Ã $ Ƈ :?Ã:;Ã Ƈ Z$@ Ƈu£Ƈ:?-5?:Ã`£5.:8 &5Ã
5\mQ@uqyM 0@~UMh@~\G@d 5qG\M~ 6]nRAvrzN1AVNiA]HAe3eiv]AK 613 B22V " " " " !" " " "
"
" "" 5 ";AD1?M"((M K± >C1AB5<;A 9) *4C1AB5<;M.,??51AM 5*± :,?8 )<M .,9.C9,B
G
¢vo¢ q {
u¢tuo¢ EN¶ ]^¶ RPµ¶ EN¶D¶ i¢ PRD¶ à g
m¢f4¢ i¢ So¢^¢ ho¢wo¢gog q vo¢ go{¢ENPR.¶ _o¢ vo¢gog¢ q¢ =NdP¶ {¢ # C{
m¢ vo¢go¢ ¢q f}i wÃ
£Ã
lÃ
rÃ
B{gog¢{g
uo¢ENT¶ g
m¢NPW¶ go¢ ookom¢ oo
g¢
¢ wo¢ {mo¢EN¶ g
m NP¶ ¢q g¢ggoug¢ENPR.¶ [o¢vg¢=RUW¶ {¢o{gog¢ So¢# jo¢ g¢ {{o¢ {
ouo¢ g
m¢~± hp¢ vo¢ uogo¢l
¢ m{{¢ ¢q # ± M g
m ##¢M ± # C{
m¢g¢vo¢{ho¢go¢q¢~± J{q ¢ ¢g
o avo¢m{soo
ko¢ hooo
¢ vo¢ mk¢ g
n¢ vo¢ ¢ q¢¢ m{soo
¢ {
ouo¢{¢ og ¢ wo¢ ¢ ¢q vo{¢D?@¢ uogo¢ k
¢ m{{¢ g
m¢S?U¢ og¢ k
¢ {o ¢C{
m¢g¢voo¢g{¢q¢
ho¢ K{q ¢ ¢g
o¢ , ¢ C¢g
¢ {{o¢{
ouo¢# o¢# mo
o¢wo¢# {{o¢
go¢{
ouo¢ { o¢ #5¢# #5¢# #5¢,¢ #5¢-¢ ok¢ [o¢wg¢ #### #¢¢k³¶ voo¢¢¸³¶ mo¡o¢ vo¢ {
ouo¢ ko¢ ¢¸$à C¢ ogo¢ ¢i³¶5¢ M ¢_³¶D¶
¢³¶D¶ # ¢6±¶ # AU
M
5\mQ@uqyM 0@~UMh@~\G@d 5qG\M~ 6]nRAvrzN1AVNiA]HAe3eiv]AK 613 B22V " " " " "
ec{v¢ ¢ ¢q uo
og{¢ o¢^¢5¢ # `vo
¢qNdZw¶# # ¢ _{
ko¢rEORx¶5¢rENQx o¢vgo¢qEdRw¶5¢rOdZw¶5¢# ¢ So¢rRdQw ¢ g
m¢qEdNw¶# ¢ `vo
¢#¢5¢# { o¢ ¢# *¢ :¢¢± ¢# ,¢ `v¢, ¢5¢* ¢ _{
u ¢ o¢uo¢# ¢ ¢ g
m5¢* _{
ko¢=RdQ¶ =NdE¶ o¢vgo¢}¢# f$¢}i$¢ `v¢fi¢# # ?U
JU
vÃ
# ¢ do¢vgo¢
¢Ã
`TOX¶ #*¢`EOQ¶5¢#*¢ 0¢`OQR¶ 5¢-¢¢`NQR¶5¢`RQW¶ @U
:¢WV¶# XQ¶ g
m¢NT¶# OE¶5¢QR/¶ `v¢=WNT¶ >XQR/¶ `vooro XT¶ X¶R/¶ _{{g¢=T¶ER¶ =RQ¶X/¶ `vooro¢TR¶5¢R¶W0¶ `v¢=R¶T¶W¶ { o{gog¢ JU
AU
IU
{Ã
}Ã
vÃ
C¢¢# ¢ g
m¢¢# # ¢ vo¢ ukm¢ go¢ ¢ g
m¢, ¢ ook{o¢ :
¢ k
¢ m{{ ~± q &¢ ± ¢ g
m¢¢± $¢ ± ¢ m{{mo¢ vo{¢ m{soo
ko¢ #¢¢ ¢ Eo
ko¢~± m{{mo *%¢¢ #¢± # ¢# ,¢ `v¢vo¢{ho¢go¢go¢ ¢ g
m¢,
-¢
*¢
9V~ ~\V ]o~V[Vx| OV MoT MoT M||lV ~\M~
jÃ
.]x|~
V os~V ~\M~ MoT MxV
Os~\ osoVxs MoT ~\M~ ~\V]x 0*, MoT 9+; MxV Os~\ us|]~]V O TVZo]~]so 9V~ 2 OV ~\V 0+, ~\Vo fg
>U
<N MoT
[l\¶"=^P
\VxV N MoT P MxV Rsux]lV ]o~V[Vx|
E\| ~\V 9*; sX MoT ]| <NQ 4X" ~\Vo <# MoT ]~| VM|]h R\VRdVT # < < E\| NP ~\M~ ~\VxV ]| os |sh~]so I\Vo jà jà 'X~Vx |O|~]~~]o[ Xsx MoT MoT |]luh]X]o[
V \MV
NP< " N P
]Ã < * ¢
YZÃ
5X <" ~\Vo NP N P# ]V N P # E\| N# P# sx ".¢ "* ¢ D]l]hMxh
\Vo 3
>U
V [V~ NP N P" ;h~]uh]o[ ~\xs[\s~
O MoT ~\Vo XMR~sx]V
V [V~ N P #
\]R\ []V| # .¢M oT" Lb\Vo <# *¢
V [V~ " "*¢
\]R\ ]| xVcVR~VT M| MoT MxV T]|~]oR~ >V~
V Rso|]TVx ~\V RM|V
jÃ
¢jÃ
Mx[lVo~|
V [V~
>U
E\Vo #
\]R\
<
]VhT| 3 I\Vo <"
V [V~ N" P E\| ~\V |sh~]so| MxV P" N$ sx # "
\VxV ©¶S¶c+¶ I\Vo # ~\V VwM~]so |]luh]ZV| ~s N P "¢ E\|
V [V~ N" MoU P MxO]~xMx M| ~\V soh |sh~]so E\V |sh~]so| MxV " "
\VxV S¶c*¶ .]oMhh
V Rso|]TVx ~\V RM|V
>U
¢ jà à jÃ
1VxV "±N "
<
jÃ
< ~\VxV ]| os
|sh~]so E\| ~\V |sh~]so| MxV
. ¢ . ¢*¢ ± ©¶ MoT
\VxV ©Sc)¶
, ¢ Fr¢#$ # e ~\Vo ~\VxW MxV VMR~h e |wMxV olOVx| hV|| ~\Mo # e ]Ã # ]Ã e >s
V |\s
~\M~ e# °¶±-¶ IV \MV e ### # e #e 1VoRV
% e
E\VxVXsxV
|s ~\M~ °¶²¶
*
>U
T ]Ã # ]Ã e e e LM =
e >U
]Ã
e
\Ã
NO
e#
e
? ± 4 T e LM
%
*¢
E\|
4[lP?tpxL/?}TLg?}[F?c4pF[L} # !!# "# # @00T¿ # !#
}¶±»¿A]¿s¼¿A11U¿
1]N1 8A11¿±¿
!!#
"(,'.. MR¿ &*()$"(. ")'. -$*'. "(,'(. $".). "(,'. ().%'$+. $'. ). !* )% . $. &*()$"(. ")'. -$*'. "(,'(. ". ). "(,'. (). -. (". ). * (. $")"". ). ))'(. . . i¿ j¿ $'. . $''(%$"". )$. ).$'')."(,'(. $'. ). $)'. ($'). &*()$"(. ,'). -$*'. "(,'(. ". "(,'. ().". (. ). %%'$%').* (. $,.-$*'. "(,'( . $.()%(.'.". )$.*(). -$*'. #(,'( . .&*()$". ''(. 7¿ !'. $. * )$'(.'. $,.
# # # # # ## # # ## #
6Z¿
rµ¡³« ¿h©¿xµ°³©§°¿
#J
#J @6$JƇA6#JƇ?JƇ # £J
F Ƈ ƇƇ!JƇ Ƈ Ƈ!JƇ Ƈm! Ƈ Ƈ )Ƈ f¿
g¿ [Ã l¿
#Ƈ
@¶
#!J 6L¿
)Ƈ$Ƈ! $ Ƈ»Ã Ƈ$ 9Ƈ ƇƇ )ƇƇƇ Ƈ$Ƈ$ !ƇJ!Ƈ Ƈ $Ƈ$ ƇƇ!!Ƈ ! Ƈ
ƇƇ Ƈm$ Ƈ !Ƈ Ƈ$Ƈ! Ƈ Ƈ $
! $ƇƇ!JƇ Ƈ$Ƈ!!Ƈ )Ƈ! ƇƇƇ RƇ$ !/*Ƈ $ Ƈ F Ƈ Ƈm!$*Ƈ $Ƈ Ƈ! ƇƇƇ Ƈ
#!J
)Ƈ f¿
g¿ [Ã
l¿
"
Q¿ S¿
(""& *J
Ƈ
½¨Ã ·½¨Ã ½Ã ·½¨Ã ½Ã ·½Ã
·½Ã ·½Ã ·½¨Ã
³ $ƇƇ Ƈ$ Ƈ Ƈ Ƈ
@@@¶´@@¶@@@¶(((¶@@@¶*# """"""""""""*$
(""+J
(""+J
W¯
N°
##-J
Ƈ Ƈ*Ƈ )Ƈ 8£ f¿
g¿ "Ƈ S¿ [Ã \! Ƈ!J Ƈ$!¡Ƈ k¿
!J
#@
WAƇ
]=Ƈ ¿ Ƈm± Ƈn± Ƈ(Ƈ %Õ EƇƇ(ƇƇE EƇƇ%Ƈ);3OƇ qEƇ % Ƈ Ƈ(Ƈ ƇE Ƈmn± ](Ƈ% ƇƇ(ƇƇƇ%Ƈ);3OƇ EƇ Ƈq ]Ƈ ±
^±
)Ƈ ;Ƈ 3Ƈ OƇ l¿
|±
, W, , Z±
,_
\±
±
FƇ EƇƇ Ƈ(Ƈ·ƇŘŪ¢Ƈŧ EƇƇ(Ƈ EƇƇƇ ? ƇEƇE( AƇ ° EƇj¯Ƈ, WƇ¶=Ƈ±Ƈc¿Ƈ¶=ƇjƇ, +Ƈ¶Ƈ ¯jƇ ±jƇ +n_Ƈ F Ƈ ¯± =U
)Ƈ ;Ƈ
nƇ ,nƇ
OƇ
nƇ ,nƇ
l¿
,nƇ
3Ƈ
f±
I)± >)± M)± ±
M)± M)±
±
+_
]( Ƈ2Ƈ EƇƇE%ƇƇ% Ƈ Ƈc¢E¿Ƈ·ƇcƇEƇƇ%EƇ = (ƇEƇƇ ƇƇ,Ƈ DƇ *Ƈ%EEƇEƇƇcƇƇ( W )Ƈ + ;Ƈ 3Ƈ 7 OƇ l¿
2_Ƈ
>%%EƇ$, EƇ(ƇEƇƇ EƇŷAƇ FƇ¢%=Ƈ,, W<&ƇMÃ , W_Ƈ F Ƈ (ƇƇƇ ,Ƈ,gƇMÃ Ã 'AAAA((__ƇgƇ'& )Ƈ ,7<Ƈ ,<<Ƈ ;Ƈ 3¿ ,<Ƈ O¿ ,<Ƈ ,<Ƈ l¿
Ƈ
7A
U(Ƈ ƇƇ(ƇƇ? Ƈ);3OƇ(Ƈ);Ƈ, &Ƈ;3Ƈ +&Ƈ3OƇc¿7 ƇO)Ƈ, AƇ U(Ƈ Ƈ)3Ƈ Ƈ;OƇ ƇƇ(Ƈ% Ƈ+¿ Ƈ(Ƈļ3P;Ƈc W,:AƇ Ƈ(ƇƇƇ(Ƈ? Ƈ);3O#Ƈ ;U
{±
)Ƈ ;Ƈ 3Ƈ OƇ k¿
<_
7Ƈ L<Ƈ Ƈ Ƈ Ƈ
{ Ƈ(Ƈ Ƈ vƇ8epIJƇƇƇ?Ƈ(ƇjƇ8 Ƈ8je
;U
-&ƇjeƇc¿-Ƈ ƇjpƇ , -#Ƈ
]±
£
eƇ )Ƈ ;Ƈ 3Ƈ OƇ k¿ _
:Ƈ :Ƈ ,:Ƈ W:Ƈ W,:Ƈ
](ƇƇ(ƇƇ%Ƈ%Ƈ¡ƇƇN 56NƇƇ2 &Ƈ(ƇNƇ Ƈ¡Ƈ %¡Ƈ < )Ƈ 2 ;Ƈ 3Ƈ 2 OƇ 2 k¿
Ƈ
|©¬³¿xµ°³©§°¿ 44
>%%Ƈ Ƈ5´MVƇ -}ƇlƇMƇ ´VƇ C}Ƈc¿VƇ ´ M }Ƈc¿Ƈ F 5 5Ƈ ƇƇƇ-ƇƇCƇƇ#Ƈ
4
F Ƈ Ƈ Ƈ ƇƇ4Ƈ
4ñ Ƈ
Ƈ '¶ Ƈ Ƈ±Ƈ ± bà 4Ƈ Ƈ±Ê,&Ƈs Ƈ ƇRƇƇƇ5Ƈ
4W
>%%Ƈ Ƈ4WŻƇc¿#ęWêŒƇc¿4Ƈ F Ƈ ƇƇ
K,_
F Ƈ ƇƇƇ K ƇBHƇ Ƈ4HƇ Ƈ4+HƇ ƇH é
4+
U ƇƇƇ Ƈ ƇÏ)UDzÏ)U
3>ƇƇŦ Ƈ Ƈ ƇƇ£Ƈ Ƈ śƇƇƇƇ Ƈ ƇƇĵƇ_Ƈ F Ƈ ƇƇ Ƈ
Ƈ Ƈ Ƈ ƇƇ Ƈ Ƈ Ƈ£_ \`ƇU ƇƇƇz Ƈ ƇƇ)&Ƈz&Ƈ
òƇ+¿~
42 Ƈ
ƇƇ >ÊƇ4&Ƈ&Ƈ&Ƈ# # &Ƈ4<<& Ƈ ƇAƇ U ƇƇ)ËƇ -&Ƈ C&ƇƇƇ > Ƈ ƇƇ ¿ Ƈ¥
Ƈ -gƇƇc¿CƇDƇ ƀƇ¥
¿Ƈ Ƈ> Ƈ \`Ƈ U Ƈ ƇƇ Ƈ Ƈ Ƈ ƇƇ Ƈ Ƈ#ƇFƇR%&ƇƇ Ƈ -¾Ƈ C§Ƈ ƇƇ -¾Ƈ§Ƈ CƇƇ ƇóƇ C§Ƈ -ƇƇƇ ƇƇ_Ƈ
47
F Ƈ Ƈ Ƈ Ƈ ƇƇ4ƇƇ Ƈ£Ƈ
4<
Ƈ Ƈ Ƈ Ƈ ƇƇă Ƈ ƇƇ,7Ƈ Ƈ ƇƇ
ƇƇ ƇƇƇ ƇƇ ¹ƇƇ Ƈ&Ƈ s Ƈ ƇƇ ĉ
zƇB+Ƈ Ƈ4»Ƈ Ƈ»Ƈ Ƈ,:Ƈ Ƈ2 »
4
F Ƈ ƇƇ ƇƇ ƇƇƇƇ ƇƇ¶
Z#RƇ4
Ƈ
3J
z
DU
;<
-Ƈ CƇ
,
, ,K6) , W6) #Ƈ
#Ƈ
>%%Ƈ Ƈ), , Ƈ), Ƈ Ƈ ƇƇƇKK6) F Ƈ Ƈ/ƇƇ ), ,) ,,),
#Ƈ
Ƈpfg¿ngef¿, HƇ Ƈoefg¿c¿7HƇ U Ƈ% Ƈq¿/Ƈ Ƈ Ƈ /Ƈ Ƈ Ƈnqeg¿, KHƇ Ƈ nqge¿c¿H #ƇF Ƈofqg¿ Ƈ Ƈ
W#Ƈ
DƇ *Ƈ%Ƈ Ƈ¬Ã /Ƈ Ƈ2Ƈ ƇƇs Ƈ Ƈ Ƈ
&U
%!% % % %"% ,
,
dà ¬Ã
Ƈ p¶)v¶ Ƈ ƇƇ Ƈ/Ƈ ƇƇ/ƇƇ),
FƇR%/&Ƈ#B}Ƈ Ƈ B}Ƈ , BƇ À#B}Ƈ , À ƇAƇ :U
B#Ƈ
ƇÐep&Ƈ/ 8eƇb !, epƇ
8pƇ , Ë
CƇ >%%Ƈ
,
,
6ą ƇKvƇs Ƈ Ƈ
!,
!,
/ƇƇ ƇƇ /ƇƇÐepƇ Ƈ Ƈ
+#Ƈ
F Ƈ Ƈ ƇƇ żƇc±ŗƁŤŴŝƇ± ±o±z}P± 6¿`¿t¿`¿ ;((OV± v}± Ƈ Ƈt¿Ƈ /Ƈ*Ƈ 2(± ± 3<± æŵƇ ¨² Ƈ
2 #Ƈ
>%%Ƈ , ƇCƇƇ ƇƇƇ)
, ,), ƇÃ ,KƇ
, ,), Ƈ Ã
Ƈ Ƈ)
Ƈ /Ƈ!, ƇŕƇŽŞƇƇƇ KƇ KƇ KƇ KƇ AƇ Ƈ, #Ƈ F Ƈ Ƈ/ƇƇ ,,+T , ,, +T :U
,
7 #Ƈ
e¿ Ƈ Ù-ųƇƇ s Ƈ*Ƈ v DÃ b¿&Ƈ %,
K ,%
,
KƇ %
!,
"
¿ Ƈ¢± #ƇF Ƈ Ƈ/Ƈ
67Ƈ
<#
ĽƇ), ± Ƈ±Ƈ·Ƈ/Ƈ Ƈ Ƈ Ƈ%± Ã ± ,»± b¿W#Ƈ F Ƈ Ƈ/ , %/Ƈ/ƇƇ) ,¶,&
ĝ#
q Ƈ ƇƇ Ù4=Ƈ&Ƈ&Ƈ# # # &ƇƇKW&ƇKBƇ#Ƈ
Ƈ ed¿ * , , +, ƇƇƇƇ qƇ Ƈ , Ƈ Ƈ , ,+Ƈ`¿ ,,Ƈ U± , DƇ *Ƈ ƇƇƇ Ƈ , 4 , qƇ
Ƈ
®©¦Ã
IJ"¿
`b
ªÃ
¸¿
1H±
.5Ã
!"#$%U
f} £¶a}¨}¨}¶f¨®¶ |§«ª¬¿r³£³ ¿u º£«¿|ru¿ :''N± |§ª¬¿|³ª§¿|ª µ³ª§¿ 4#
)
4Ƈ ¸ >4Ƈ ¸>4Ƈ ¸ J KƇ ,
GÃ
±
Ã
NÃ
PÃ
U
,
"Ã
G>>SÃ
,
H>>UÃ
'Ƈ ±FT ±ET
±ƇG>>VÃ bƇ G>>VÃ bƇ4Ƈ"ƇÃ GÃ
NÃ
PÃ
G>>UÃ
GÃ
Ƈ(ƇƇJƇ(Ƈ ƇƇBƇ L
3Ƈ q/£'Ƈ
RT
µ¼Ã
à µ¼Ã ª >
Ã
RT
µ¼Ã
%
¼Ã =à C µ¼B=Ã
&Ƈ((Ƈ % Ƈ%/sŢ
*/ Ƈ3ƇƇ(Ƈ
) PƇ(Ƈ(ƇJƇ(ƇƇ ƇƇ /Ƈ*Ƈ44&Ƈ( Ƈ(Ƈ JJ Ƈ /Ƈ /Ƈ*ƇKƇ D Ƈ(Ƈ ƇƇƄG
"AƇ
O q///Ƈ] ŇʼnƇƇ(/JƇJƇİ;3OAƇ U ũƇŀņ ]ƇƇ(/JƇJ ] ňŊƇ U(& q ]ƇƇƇ?ƇJƇy;3OƇ D ƇO
BG
z
=)±
±
=)±
M)±
±
]ƇR Ƈ(Ƈ ƇsţƇƇ(Ƈ=Ƈ Ƈ Ƈ?//Ƈ /Ƈ{ƇƇ Ƈ %/ƇƇ(Ƈ(Ƈ nvwy¿ dà h n ƇD vƇ nwy{¿bƇnƇ,KLnAƇ +
) L2Ă BƇ'ƇLLƇ t¿ƇƇJƇJ LLƇ ƇLLƇlƇLƇRƇ2Ƈ RƇ44ƇRƇ4#
LBƇ
I(ƇƇ/(ƇLRLRLRLƇ , +ƇJƇJƇd± DvƇu£ƇR Ƈ,GƇ >v u£ ƇƇƇƇL#Ƈ D Ƈ(ƇƇ"Ƈ%Ƈ(ƇJƇd±
Ĵ(vƇ-ēƇƇ-5ƇƇ-MƇƇ-Ƈb¿G Ƈ #Ƈ \Ƈ(Ƈ2Ƈ , "ƇRƇ,Ƈ^ D Ƈ(ƇƇ?Ƈ,ƇRƇƇƇ2ƇƇ"<ƇƇ"Ƈ ,<
7G
{±
O
ıƇJƇ? //ÉƇ Ō . tƇ9 tƇ Ƈ 9 u Ƈí. uƇ Ƈ", :G Ƈ r Ƈ3 Ƈ°vƇƇ(Ƈ . 5Ƈ t5Ƈ . t ", :Ƈ b 5 ðƇ t5ƇƇ9 5Ƈ ú t9ƇƇ",:Ƈ a¿+5Ƈ 9 5Ƈ^Ƈu5Ƈ^Ƈ 9 u ",:Ƈ XƇ7 5Ƈ . 5Ƈ^Ƈu5 T. u ",:Ƈ XƇ5Ƈ
Ƈ Ƈ Ƈ "Ƈ
6L¿Ƈ6"oƇ . t^Ƈ9 t^Ƈ9 u^u. ",:Ƈ X 5Ƈ6+ 5Ƈ^7 5ƇT 5Ƈ T+^+"6"Ƈ #Ƈ bƇ#Ƈ D ƇƇJƇ? ŲƇ, "Ƈ
<#
3 ĻƇ8eƇ ,.Ƈ ƇjƇ8eƇ, ~GƇ ;*Ƈ Ƈ3 Ƈ°/Ƈ ƇŁ*(äƇI(Ƈ ( . 5Ƈ^Ƈ- 5ƇT-. ~ƇƇ- 5Ƈ .T- ~ 5Ƈ .6- 1 ~ 5Ƈ X -5 #Ƈ ]Ƈ ƇŜƇ(ƇƇƇ? Ƈ(Ƈ
L+Ƈ
.0ƇT-0Ƈ
-.Ƈ r Ƈ0Ƈ~Ƈ0~Ƈ'Ƈ=ƇƇ Ƈ
-.Ƈ
.0T-0Ƈ 9J .0T,-0Ƈ 9JbƇ"-0.0 Ƈ Ƈ Ã
I . 6-0.0ƇƇ2 Ƈ > =ƇƇ Ƈ.ƂƇbƇ ,Ƈ± © 0 = ƇƇ Ƈ.Ƈ į-&Ƈ Ƈ Ƈ .0Ƈ XƇ ,ƇƇ © 0 9U
D ƇƇ8jeƇ'Ƈ û
© =ƇƇ Ƈ? Ƈ ƇƇ,: Ƈ
z \Ƈ Ƈ¥?NƇƇ2 ¥[¥Ƈ FƇ Ƈ ƇƇƇƇ%&Ƈ Ƈ¥¿'Ƈ =Ƈ Ƈ ƇƇ¥?NƇƇ2 2 ƇƇ¥¿'Ƈ&Ƈ Ƈ ƇƇ¥?NƇƇ2Ƈ'Ƈ2 Ƈ > ƇƇ%Ƈ Ƈ ƇƇƃ Ƈ Ƈ Ƈ*&ƇƇ Ƈ Ƈ2Ƈ Ƈ *Ƈ ;U
8U
y xƇ Y[ MƇ Ƈ-}Ƈ'ƇƇ %Ƈ MƇƇ-lƇƇ D ƇƇ-'Ƈ=Ƈ Ƈ-'Ƈ+" 0Ƈ >¹*=Ƈ CƇ'Ƈ+Ƈ Ƈ
¿'Ƈ[$¿ I ŶƇ-Ƈ ± CƇ ±
¿ b¿Y[%¿
y xƇ 2dÅƇ/T 2 dÅƇ/T 2Ƈ Ƈ Ƈ2Ƈ ´Ã <Ƈ´Ã 2Ƈ'Ƈ""Ƈ.T ƇœƇç&Ƈ Ƈ Ƈ ÔƇƇ
) xƇ ""Ƈ rƇ Ƈ *Ƈ.0ƇƇĊƇ'Ƈ.ƇƇ90TſƇ ""6Ƈ'Ƈ"" \xƇ { Ƈ Ƈ&Ƈ Ƈ*Ƈ ƇƇƇƇÔ Ƈ Ƈ Ƈ*Ƈ¨ ¨ Ƈ ƇƇƇ.Ƈ Ŕ9ƇƇ Ƈ ƇƇ Ƈ ;U
"
ÌxƇ ] Ƈ
Ƈ ĔČƇ"
Ƈ CƇXƇ
Ƈ ĕčƇ"T"Ƈ
* Ƈ {| y¶XƇ -Ƈ CƇ ,
) ġƇ , Ƈ|| Ƈ,HƇ ƇHƇ Ƈ+HƇ ƇH 'Ƈ Ƈ2 HƇƇ"2H6 Ƈ2HƇ Ƈ"HƇ 'ƇƇƇ Ƈ"Ƈ2 H Ƈ Ƈ2HƇ6Ƈ"Ƈ2 H Ƈ Ƈ2HƇ 'ƇƇƇ"2 H62 HƇ'ƇƇ ƇHƇ ' ,Ƈ
+
) xƇ ,Ƈ 2Ƈ
D Ƈ Ƈ% Ƈ
"áƇ 2 ĖƇ I 1 11ÉƇ ōŎºŋƇƆ b º+ b ,G ¼Ƈ ¼ÛƇ Ƈ2 # Ƈ
yŰoƇ <<Ƈ >1ƇÃƇµƇƇ1&Ƈ Ã 1 ƇµƇ Ƈ ƇƇ%*Ƈ Ƈ Ƈ
ƇƇ Ƈ1GƇ I ƇƇƇ
Ƈ1 ƇƇ1Ƈ1Ƈ1 >#ƇP1ƇÃ 1 ƇµƇ 1&ƇCƇƇ
:;=JbƇƇ<<ƇbƇ<<
GƇI Ƈ1ƇƇ¥
¿ƇƇ Ƈ 7 ĢƇ
)1oƇ Ƈ \Ƈ Ƈ»EEÃƇƇƇ Ƈ*Ƈ»ÃƇƇ*ƇFƇI ÁƇ D1Ƈ ƇƇƇ'Ƈ ėĘƇ ½ ƇƇ Ƈ*ƇƇƇƇ'Ƈ&Ƈ#ÁƇ ƇƇƇƇ Ƈ*Ƈ#Ƈ
<#
)1oƇ ÍƇ Ƈ1ƇƇ¥¿1 Ƈ ¢¿ Ƈ¥¿ià ¢$ ¥¢¿' ,7 ¥>¢>¿'Ƈ¥¢¥¿½Ƈ¢¶lƇ,7NƇ îƇ¢¶ƇƇ%ƇƇ&Ƈ Ƈ& ,7NƇƇ¢¶' ||Ƈ ƇƇ1ƇGƇ I Ƈ%ŖšƇƇ <NƇ Ƈ¢¶'Ƈ,Ƈ >1Ƈ <&Ƈ,'Ƈ&ƇƇƇ Ƈ ¥¿Ƈ¢¿' Ń |ÖƇƇƇ%Ƈ1ƇÖG >1Ƈ ƇƇ1*Ƈ Ƈ¥¿,-T ¢¿l ,7Ƈ1 Ƈ1Ƈ1Ƈ 1 Ƈ Ƈ
¢¿1 Ƈ¥¿ƇƇ Ƈ1&Ƈ1*Ƈ¥¿'Ƈ2 < Ƈ1 Ƈ¢¿lƇƇ Ƈ Ƈ?1G I Ƈ ƇƇ1ƇƇ#
G
)1oƇ + ,+Ƈ1Ƈ :Ƈ1Ƈ:Ƈ1Ƈ ,:Ƈ1Ƈ2 : â Ƈ" :ƇƇ :Ƈ ƇºƇ,+Ƈ1Ƈ :ƇƇ :Ƈ1ƇĎƇ :Ƈ 7 1 :Ƈ :Ƈ " :Ƈ :Ƈ +"1 " :Ƈ " :Ƈ :Ƈ 1 7 :Ƈ :Ƈ +#Ƈ
#
ÌoƇ , r1Ƈ1Ƈ ƇƇƇ*ƇƇƇ
Y 8J TY 8JX ,& 1 Ƈ Ƈ 6Y¶ &Ƈ Ƈ ¬ TY 8J¬Ƈ ×Ƈ¬Ƈ Y 8JƇ×G 7Ƈ
G
) xƇ + ÍƇ-Ƈ'ƇƇ6¶Ã ƇCƇ'ƇƇ6¶#à D Ƈ Ƈ? Ƈ -MƇ ¦ƇCMƇ -ƇƇCƇMƇ ƇƇ r Ƈ Ƈ ƇR% Ƈ -ƇCß'Ƈ-ƇMƇƇCƇMƇ ¦Ƈ-CƇ- ƇC&Ƈ z? ƇƇ Ƈ -C - C' |Ƈ I Ƈ Ƈ ƇƇƇ Ƈ -Ƈ'Ƈ&Ƈ CƇ'ƇƇ Ƈ-ƇƇC Ƈ Ƈ? *&Ƈ Ƈ ƇƇ Ƈ Ƈ? ƇƇ¶Ã'Ƈ&Ƈ ƇƇG D Ƈ ³AÃƇ¶ JÃƇ¶Mà Ƈ¦ƇƇƇ Ƈ +GƇ 5U
)U
6U
3U
A
2U
yűģƇ :G 3 ƇƇƇƇ Ƈ Ƈ);3&Ƈ Ƈ*¿Ƈ Ƈ ƇƇ G
-UU,U
\Ƈ Ƈ[)3;Ƈ'ƇğSƇ > Ƈ P3Ƈ P;Ƈ Ƈ[ 3P;Ƈ'Ƈ+SƇfà [P3;Ƈ'Ƈ+SƇ Ú3P;ƇƇ Ƈ?Ƈ GƇ I Ƈ[P3)Ƈ'Ƈ2 S6+SƇ 'ƇSƇ'Ƈ[P)3Ƈ ;Ƈ[i)3Ƈ'ƇSƇ ÂƇ >Ƈ)PiƇƇ ƇƇ ƇÕ #Ƈ [)P3Ƈ +SƇeà [3PiƇ'ƇS#Ƈ > Ƈ[i3)Ƈ'ƇSƇ Ƈ Ƈ[P3)Ƈ'Ƈƅ&Ƈ Ƈ[i3PƇ SAƇI Ƈ Ƈ Úi3PƇƇ ƇƇ Ƈ Ƈ[i3;Ƈ 2 üƇ "SƇ >Ƈ;iƇƇ Ƈ%% ƇƇƇ Ƈ?Ƈ Ƈ P;3#ƇI Ƈ [P;iƇ'Ƈ+SĆ Ƈ'Ƈ SƇ [;i3Ƈ'Ƈ7S6"SýSƇ 'ƇďSƇ 4U
4U
4U
5U
"#
) ĤƇ "
<Ƈ
7U
7¶% ;¶ ¶ & % % %' %% %%( %'
$¿ ķŅ Ƈ >Ƈ
¦
¦
¦
¦¿ ¦¿ ¦ ¦¿ ƇƇ Ƈ 8¶ ;¶ ¶ ¦
'
XƇ
T
T
I Ƈ¥¿ƇƇƇ%ƇƇ+Ƈ w I ƇƇ "Ƈ Ƈ #
,#
) oƇ :# {ƇƇ Ƈ
E = C E
ÂƇ I Ƈ ƇƇ Ƈ Ƈ
'C = E ( !=C
;*Ƈ Ƈ ƇƇƇ &Ƈ ƇpƇ ƇƇƇ ƇƇ #
+#
C = E
1U
=C
Ƈ D ƇpƇlƇ: Ƈ
) ÇƇ " \ƇƇ Ƈ Ƈ*ƇƇ
%
&Ƈ '+wƇ
\ƇƇ Ƈ Ƈ*ƇƇ
\ƇƇ Ƈ Ƈ*Ƈ ƇƇ ƇƇ ãƇ
% w
XƇ# +Ƈ ;*Ƈ Ƈ% %ƇƇ Ƈ Ƈ¢ &Ƈ Ƈ ƇƇ Ƈ Ƈ*Ƈ ƇƇƇƇƇ lƇƇ +w6ƇlƇ"#
w#
y oƇ ]Ƈ Ƈ )&Ƈ
=C
EJ 6Ĉ 9U
= &Ƈ Ƈ C' Ƈ C = èƇ Ƈ D Ƈ J & Ƈ E' E J.
# D Ƈ
TU
U
= C) !=C EJ !EJ Ƈ Ƈ Ƈ> Ã
Ƈ
Ƈ
L7AƇ
) `Ƈ "Ƈ Ƈ AƇ > Ƈu-± , Ƈ u , ¨ƇàƇuMƇ , uV 0Ƈ ¤Ƈ ¤Ƈ ƇƇu 9± ƇƇ Ƈ u¡ . u 9± , u± G± Ƈ Ƈu9± T-ŮìěƇ Ƈpq± 6u 9
BU
GU
Ƈ
0
¤Ƈ
uJ±
2U
= ƇƇ ƇƇƇ đƇÜ ƇƇ
'U
u
XƇ
.u±
«Ã c ± Ã
-Ƈ ö÷ÝƇ 6-Ƈ±
, , 0;;TÃ uV (R; #, TMƇ uD±
u
Ƈ
õ 7Ƈu & ;@ , " 0 <ćƇ
) `Ƈ "Ƈ rƇ Ƈ *Ƈ
FJ & Ƈ Ƈ
.0Ƈ ± 90Ƈ 0Ƈ6.9 Ƈ9 . ƇÈū .69 0Ƈ96 0Ƈ.6 0Ƈ Ƈ Ƈ Ŭŭ *AƇ AƇ
) `Ƈ +,Ƈ >%% Ƈ ƇƇ.dƇ ƇŹƇ Ƈ-d&Ƈ. Ƈ Ƈ-dƇ Ƈ- &Ƈ M 0Ƈ 0Ƈ Ƈ Ƈ- Ƈ-ÄƇ Ƈ VƇ ƇƇ Ƈ-MƇAƇ 0Ƈ F Ƈ Ƈ ƇƇ%ƇƇƇY± Ƈ? ƇƇŚ Ƈ Ƈ Ƈ Ƈ ƇƇ Ƈ? Ƈ .dƇƇ. ±.ÄƇ Ã.VƇ Ƈ Ƈ Ƈ Ƈ Ƈ 0Ƈ ĹƇ Ƈ&Ƈ ± &Ƈ MƇ Ƈ Ƈ VƇ AƇ 0Ƈ I Ƈ ƇƇ ƇƇ ƇƇ? ƇƇ? ŏ ƇƇ Ƈ ƇƇ ƇƇ Ƈ? 9dƇ ± ÃĜƇƇ VƇ , 7&Ƈ Ƈ9d&Ƈ9 &Ƈ MƇ 9VƇ 0 0Ƈ ů Ƈ AƇ
FJ
FJ
D Ƈ Ƈ ƇƇ AƇ
0U
FJ
2<J
, +,A
) `Ƈ "Ƈ ;*Ƈ Ƈ &Ƈ ƇƇ Ƈ VƇ V .ƇØƇ .6ØƇ 6T . +.090Ƈ ëƇ 9 I Ƈ ƇsƇ? ƇƇ V VƇ . Ƈ+.0Ƈ90Ƈ Ã 9 "Ƈ VƇ VƇ . Ƈ+.090Ƈ ± 9 ÈƇ,+ 2U
.0ƇT90Ƈ0Ƈ
±7.0Ƈ90Ƈ XƇ,+
+0Ƈ ±7.0Ƈ90
2U
,+Ƈ
Ƈ
U =Ƈ
. 5 ^9 5 5Ƈ . 569 5 5Ƈg ". 59 5Ƈ X B+Ƈ
D Ƈ . 5Ƈ^Ƈ9 5Ƈ
y `Ƈ B 78Ƈ > Ƈ 78Ƈ
XƇ"Ƈ
=U
=U
Ƈ78ƇXƇ
8T 8Ƈ #Ƈ #Ƈ =Ƈ *Ƈ YYƇƇ Ƈ8 ^Ƈ Ƈ8Ƈ
Ƈ78ƇƇ8g Ƈ Ƈ8'ƇƇ78ƇƇ8ø Ƈ8Ƈ Ƈ78ƇƇ8gƇƇ78Ƈ Ƈ8Ƈ'ƇƇ78ƇƇ8Ƈ Ƈ 6 Ƈ78Ƈ Ƈ8Ƈ Ƈ78Ƈg8Ƈ'Ƈ Ƈ78Ƈg8Ƈ Ƈh8Ƈ'ƇƇh8Ƈ Ƈ Ƈ Ƈ Ƈh8Ƈ, Ƈ U Ƈ// Ƈ%/Ƈ h8Ƈ'Ƈ"B:=Ƈ ÞYƇ.Ƈ'ƇBƇ
) ÇƇ B { ƇƇY Ƈ ƇNƇ ƇƇ% Ƈ Y# A<<Ã
ƇNƇ 8± =Ƈ
, YN 6_a
9U
Ƈ¥ BĞ=ƇƇ Ƈ Ƈ Ƈ/ƇƇ"h BƇ Ƈ
§aAÃ
Ƈ ƇNƇ'Ƈ ĒƇ A <<Ã
ƇNƇ [Ã =Ƈ
, YN T_a
;U
N 56 Ũ|ş NgƇBBƇ { NƇ ƇƇ% Ƈ Y=Ƈ Ƈ Ƈ
§aAÃ
/Ƈ ƇƇ Ƈ ƇNƇ /ƇƇB#Ƈ
"G
:U
BƇƇNƇ'ƇBƇ /*Ƈ ;*Ƈ Ƈ Y=Ƈ Ƈ Ƈ
) `Ƈ " .ƇgƇ9Ƈ'Ƈ2Ƈ Àà .Ƈ'Ƈ2 69Ƈ'Ƈ++h 69 > ƇƇ ƇƇƇ/ /*Ƈ%=Ƈ Ƈ//Ƈ Ƈ. Ƈ /Ƈ*Ƈ#Ƈ ] Ƈ. f=Ƈ ƇfƇƇƇ% Ƈ YƇ U Ƈ? Ƈ Ƈ 9Ƈ ++hƇ ù fƇ > Ƈ++h 6fƇhà =ƇƇ ƇfƇ[à "BïƇ Ƈ Ƈf ƇƇ% Ƈ Y=ƇƇ Ƈ ƇħĨƇ fĩĪƇ "#Ƈ 3 /*=ƇƇ *Ƈ% ŠƇ YƇfƇ YƇƇīĬƇ fĭĮƇ "=Ƈ ƇƇ/*Ƈ Ƈ Ƈ f=Ƈ++hƇ6fƇƇƇ%ƇƇ% Ƈ YƇ Ƈź Ƈ ƇY Ƈ? Ƈ U Ƈ ƇƇ"Ƈ%ƇƇ% Ƈ YƇ YƇ ƇY Ƈ? #Ƈ /U
1U
Ƈ
,
'
'' ' ' ' ' ' ' '' '' % ''' ''' ' ' ' ' ' ' ' !" #" ' $' % '
&
' ' ' '
'
) `
\Ƈ(
U
FÃ
OÃ
.
Ƈ/ƇƇ
D Ƈ(Ƈ? ƇƇ ƇƇ Ƈ*Ƈ(Ƈ( ƇƇ Ƈ
SU
D Ƈ
´Ã {¿,
Ƈ
5\mQ@uqyM 0@~UMh@~\G@d 5qG\M~ 7]nRAvr{N1AWNjA]HAe3ejv]AK 613 B22V " " " " "
!" " " " " ""
";@D0?M"((M K± >C0@B5<;@ 9) *4C0@B5<;M -,??50@M 5*± :,?7 )<M-,9-C9,B
¢ 4~ ]| []Vp ~\M~ E MG MKM MxV MxVMh plOVx| |R\ ~\M~
GKM KEM ± M#EGM M GKM M± KE ±
>U
¢
4| ]~ ~xV ~\M~ M~ hVM|~ ~
s sX ~\V ~\xVV plOVx| l|~ OV VwMh&
8|~]X sx
Mp|
Vx
/sx Mp us|]~]V ]p~V[Vx # hV~ # TVps~V~\V q~\ us|]~]V psp|wMxV ]p~V[Vx ]V
2M" M 2M M 2M",¢ 2 M -¢ V~R @xsV ~\M~ >U
>U
!"#
±¶ TVps~V| ~\V ]p~V[Vx Rhs|V|~ ~s ¸$à /sx VMluhV !# "#" M !#"#" !#"#" M!#"# M
\VxV !#
FU
AU
9=> ¢ ¢ W¢ MxV ~\V m]Tus]p~| sX ~\V |]TV| 9=¢ 9> xV|uVR~]Vh E\V h]pV T¢W¢ ]p~Vx|VR~| ~\V R]xRlR]xRhV sX /9=>¢ M~ M¢ MpT b± MpT ~\V h]pV| >¢M¢ MpT >¢a± lVV~ ~\V h]pV 9=¢ M~ Y¢ MpT \ ¢ xV|uVR~]Vh @xsV ~\M~ e±9 $¢ \=¢"\9$¢ e±= 4p ~\V Vw]hM~VxMh ~x]Mp[hV
QU
MU
E\]x~ ~
s uM]x| sX ]TVp~]RMh ~
]p| MxV h]pVT u ]p Mp 0¢ ²Ã 0¢ XsxlM~]sp @xsV ~\M~ ]~ ]| us||]OhV ~s R\ss|V
MuVx|sp| spV Xxsl VMR\ uM]x sX ~
]p| |s ~\M~ ~\VxV ]| M~
hVM|~ spV R\s|Vp uVx|sp ]p VMR\ xs
MpT ]p VMR\ Rshlp
,
.]pT ~\V lM]ll MpT l]p]ll sX
EM± G
AU
EM#GM|R\ ~\M~ M± GM±
M
5\mQ@uqyM 0@~UMh@~\G@d 5qG\M~ 6]nRAvrzN1AVNiA]HAe3eiv]AK 613 B22V " " " " "
KV} ;h~_uh_o[ Os~\ }_TV} O #¢¢#¢¢#¢¢ ¢
V [V~
W¶5 1 ¢ #¢#¢¿¢¢ #¢#¢¢¢¢ #¢¢ #¢¢5¢ >s
xV[MxT W¶ M} M ushosl_Mh _o ¢ D_oRV W¶5¢¢
\Vo ¢5¢¢
W.¶ D_l_hMxh ¢¢ MoT ¢¢ MxV Mh}s XMR~sx} sX _±
¢_}M XMR~sx sX D_oRV W¶ _} sX TV[xVV '
W ~ Xsx }slV Rso}~Mo~ ~ ¢ ) hV~~_o[ ¢5¢
¢¢5¢ ¢ ¢5¢¢
V \MV ~¢5¢#¢ E\}
W #
# '
DVV 8o_sx DVR~_so CsoT # ¢BV}~_sr ,
9V~~\VxMT_} sX~\VR_xRlR_xRhV OV h±5¢*¢MoT\V SVoyV OV &± 9V~]¢ OV~\VXss~ sX ~\V uVxuVoT_RhMx Xxsl &± ~s TV¢ E\Vo X]5¢XW#¢5¢¢ M]¢5¢
$¢
@U
$¢5¢I(¢ E\} MT¢5¢M] T]¢5¢IHI(¢ D_oRV L¢ x]¢5¢ NYTM¢ NYA>¢ MoT A>¢5¢¢
V \MV YTYA¢5¢TMA>¢5b;¢ * _V YT5¢b; YA*¢ E\} Y8¢5 YT¢ T8¢5YT¢AT5¢#YT¢ YA¢5 b;¢ YA#¢ MoT A8¢5¢Y9YA¢5b<¢'YA}# ¢ E\VxVXsxV Y8$¢5¢YA¢-( D_l_hMxh \9$¢5¢\A¢ A9¢ F\} Y8$¢ \A¢5¢\8$¢ YA JU
MU
¢Ã
Ã
',¢
KU
g
¶¦«©B¶ >s~V ~\M~ O=>P¢6¢ Q>=M¢ MoT O=M>¢5¢-¢
E\| Q=>P¢ Q=>Y¢6¢ Q>=M¢ Q=>M¢7¢ 0 Q=M>¢5¢ #¢ E\]| ]luh]V| ~\M~ Q=\>¢7¢ 0Q>=\Q=>\¢6¢ #Q=>P¢6¢ Q=>Y¢ MoT~\VxVXsxV |=>\¢ |=Y>¢ %¢ 7¢ >s
|]oRV 7¢ E\| >\¢Y>6¢ =\¢=>¢5¢=>¢¢=Y¢ ]V
JED J BDJ
HU
BCJ
MP¢ 66 =>¢ 8¢ ]| Mh|s l]Tus]o~ Mhso[ ~\V MxR sX MP¢ E\| QM>8¢7¢ Q8>R¢ ) ~\V Mo[hV O]|VR~sx ~\VsxVl so |Y>\¢
V \MV >\>Y¢5¢8\¢8Y¢ *slO]o]o[ %¢ %¢ ¢ %¢ %¢
V [V~ V Y8 \= \8 \=¢
15J 05J .5J /4J 41J 04J
7
*
Duus|V so ~\V Rso~xMx ~\M~ ~\]| ]| os~ us||]OhV IV |M ~\M~ M xs
sx M Rshlo ]|
¬
¤
¶ ]X ]~ Rso~M]o| soV sX ~\V R\s|Vo uVx|so|
*\ss|V '#¢ uVx|so| |s ~\M~ ~\V
~s~MholOVxsXRsVxVTxs
| MoTRshno|]| lM]ll I ]~\s~hs||sX[VoVxMh]~ M||lV ~\M~ Rshlo
¢ ]| os~ RsVxVT
E\Vo os uM]x sX ~
]o| RMo OV ]o Rshno
E\V Rso~VxuMx~| sX ~\V|V uVx|so|l|~ OV R\s|Vo ) ~\V lM]lMh]~ RsoT]~]so ~\V|V uVx|so| l|~ OV ~\V soh R\s|Vo uVx|so| V]~\Vx ]o ~\V]x xs
sx Rshlo 4X M
¢ R\s|Vo uVx|so ~\Vo ~\VxV MxV -¢ s~\Vx uVx|so| VRhT]o[ ~\s|V ]o Rshlo ¢]o ~\M~ xs
\s MxV os~ R\s|Vo D]oRV '#¢ uVx|so| MxV os~ R\s|Vo
V \MV M~ ls|~ *¢|R\ xs
| D]l]hMxh ~\VxV MxV M~ ls|~ '¢ |R\ Rshlo| MoT
V \MV xs
Rso~M]o| soh
M Rso~xMT]R~]so OVRM|V
V \MV |R\ xs
| sx Rshlo|
9¶¦«©B¶ IVhh |\s
~\M~ M~ ls|~ ¢vVx|so| MxV oVVTVT >s~V ~\M~ ]X M |V~ sX uVsuhV RsVx Mhh ~\V xs
| MoT Rshlo| ~\Vo ~\V
]hh |~]jh \MV ~\V |MlV uxsuVx~ ]X ~
s xs
| sx ~
s Rshlo| MxV ]o~VxR\Mo[VT E\| ]X ~\V RsoZ[xM~]so sO~M]oVT O ]o~VxR\Mo[]o[ |slV uM]x| sX xs
| MoT |slV uM]x| sX Rshlo| RMo OV RsVxVT O M |V~ sX uVsuhV ~\Vo ~\V |MlV |V~ sX uVsuhV
]hh RsVx ~\V sx][]oMh RsoZ[xM~]so IV Mh|s hV~ Tz¢ TVos~V ~\V RsoZ[xM~]so sO~M]oVT O TVhV~]o[ ~\V Zx|~
¶ xs
| MoT
Rshlo| sX ~\V sx][]oMh RsoZ[xM~]sz{ :V~ | TVos~V M uVx|so O NP ]X \V ]| ]o ~\V N~\ Rshlo P~\ xs
*\ss|V
¢ ¢
¢ ¢
4o T"¢ ~\VxV ]| M uVx|so
\s ]| os~ ~\V
I]~\s~ hs|| sX [VoVxMh]~ hV~ ~\]| OV #¢#¢ DVhVR~
#¢#¢ ¢ ¢ #¢#¢ + ,+¶¶ ¶ ¶ ¶ \à -¢ MxV ~
]o| ~\Vo ]o ?a` ~\VxV ]| M uVx|so
\s ]| os~ Mlso[ ~\V]x Rso~VxuMx~| 9V~ ~\]| uVx|so OV ¢ ¶ ¢ ¶± MoT R\ss|V ~\]| uVx|so 4o ~\]|
M
V RMo R\ss|V ¶ ¶ ¶ 7¢ ¢#¢ ¢¢-¢ ,Vos~V ~\V Rso~VxuMx~| sX ~\V|V -¢ uVx|so| O J MoT ~\V Zx|~ -¢ uVx|so| sX ~\V ~\ MoT ~\ xs
| MoT Rshlo| O 8!¢ 8%¢ 8)¢ 8+¢ xV|uVR~]Vh 9V~ 6J 8y¢5¢N` E\Vo f"¢ f%¢ f)¢ f+¢ 23¢ -¢ '||lV ~\M~ f"¢ :¿ f%¢ :¿ f)¢ 9¿ f+¢ Duus|V f!¢5¢-¢ *\ss|V MoT R\ss|V M uVx|so ]o 8%¢ MoT M uVx|so ]o 8+¢
\s MxV os~ ~
]o| MoT MxV os~ ~\V Rso~VxuMx~ sX ? &¶ E\V|V # uVtuhV
]hh RsVx Mhh ~\V xs
MoT Rshno| 4X f!¢ \Ã -¢ ~\Vo f%¢ ^_¿ '¢ f)¢ 23¢ #¢MoT f+¢ 23¢ ¢ E\Vo ~\VxV V]|~A`S¶8y¢|R\ ~\M~A` J ¶5¢ ¢#¢'¢*¢MoT os~ ~
s sX AiZ%¢)¢A MxV ~
]o| *\ss|V ~\V|V *¢ uVx|so| M|
Vhh MoT ~\V ¢R\s|Vo uVsuhV
]hh RsVx Mhh ~\V xs
| MoT Rshlo|
Rso~VxuMx~ sX
4o [VoVxMh ]X os ~
s sX
'-¢
,
IV \MV
EM¢GM ¢ @U
M¢
¢M±
±
M¾¿
MGM¢
MoT
M¢ EM± GM% 9V~ % EMMGM G\Vo
V \MV
¢MM
BB M ¾¿ ¢ MoT BMBM MG¿ ¢ D]oRV
MGM¢
MMF¿
+¾¿ BM ¾¿ M¢+ E¿ MM' M ½¿ M %!M
à gÃ
BM G¿ ¢ ~\V lM]ll MhV ]| MM,MoT ~\V l]o]ll MkV ]| M± -
>s~V ~\M~ ~\V lM]ll MhV ]| M~~M]oVT
EMMGM MM%6M @U
\]hV ~\V l]o]ll ]| M~~M]oVT
h¢~\V |sh~]so sX
MoT
EM5¢
M M
h¢~\V |sh~]so sY
EMMGM MMLM >U
M
MoT
GMMM5¢¢
'1?-)BAE+Q $)G.+;)G1*)8Q 'A*1+GIQ # # # #
Z
# , &! , !) , , ,
%+, , +, , %&$)&! %, &!,! &%& &%,
#)0'5 5 \lÞ &.),$#) ',52$.'5 #)0')5#5,5#)0'5),5%'$/5 #5 )5 ,5%%($%',5.!) !$05 2$.'5 #)0') d8Þ $5 ),%)5'5#5 ,$5.),25 2$.'5 #)0')
5 &.),$#5 '')5 OÞ "'5 l7 $5 !.!,$')5 '5 !!$0 O9
\6
O - ơ Jơ ' ơơ îo$ơ*ơơơơ 3<ơơBơYơ 5ơO U-ơ Ùơơ ô *ơ$*ơơîơ* ơơơ ơ ơ*ơơƑ ơơơ*ơ5 ơ Šơ 5 * ơơ 5 $%Z Y#ơ
- Ùơ ơ *ơ ơ ơ o<ơ ơ 8* ơ ơ ơ Ãơ ơ ơO þơ ơơ ơơÿơB- Jơ 4Z ơơơơơơ vơơơ * ơ ơơ $-ơ ňơơ vơơ ' v* ơ ơ ơơơ $Ģơ ŕ*vvơơ ơ* ơ ơ ơ *ģơ Aơơ*ơơơ ơơ ơ ơơơ $$ơơ Y-ơ áơ ơơơ ơ dĐơ *$ơ *IJ O <
Þ 25 Þ \ Þ Q "#Z -ơ
B~ơ
e # ơáơơ%ơơ ơ ơeơ ơơơơơqơ 0ơơ0 ơ ơ ơơ ơ0ơơơ ơơ Ķơ ~- Jơ:>2ơơ ơ )ơ JơFơ ơIơ ơơơ ơơơ% ơ :>ơ ơ>2ơ0 ơ ơ :Fơơe F>ơ ơ>Iơ O I2#ơ +0%ơ ơ :Iơ 2Fơ %ơ ơơ¯ # ơ% o » ơ » ơơ$ơ0%ơ 0 ơ ơ:¯ơ Y¯IËơ 7Í J kơ =O5 5 BM Y5 ƞ ƟM q > Ìơ +ơBÉ$%ơ 0ơ =2 3 4>@ ơ kơ ơ ơ ơ ¯ÅÅ©Þ 2@@3@@4@ ơ ơ ơ B-ơ %ơ ơ 0%ơ ơBÉ %ơ ơ kơ ơ O #ơ Jơ <8 @< @ #@Þ °q?Þ ơ ơ ơ ơ <8@@< @ơơơ @ %ơơ ơ ơ 0ơơ Ó ;@1?@@@Ó ;sq?ÜÞ
sq?Þ
ơ e¨
OO¨ %ơơ ơ 0ơơ2@0 ơơ2@ơơ 0ơ2@ 37 02@ eơ ĵơ ơ 3@ Þ4@@37@ 2@@@ ơơơ%ơ ơ0%ơ 3@ơ4 Oĩ¨ơ Ú%ơơ0%ơ ơ ơơ%ơơ0%
' e b eơ
1BĤ Ú%ơ ơ 0%ơ ţơơ 2 @3@ơ ơ ơOơ /Z 3@ 2 2@ Z ơ 0 ơ ơ0%ơ 2 @3@2 3@23@@293@ơ ơ 0ơơơ0%) 1Ĭ- ơ 0ơơ ơ$1q- ńơ ơ 0o ơ ơ :>25ơ ơF5ơI5ơ ơ Þ ơ ơ ơ ơ ơ Ɯ 0żơ%ơ :5>5ơơ2ơ ơ>25 :2ơơ :>5ơ $)ơ µ0ơ :ơ ơ O ơ> 2 ơ8 B75ơ ơơ ơơ :ł5ơ ơơ ơơ ơ :Fơ ơ>I#
Jơ ' ơơ ơơơ 0% $ơơĺ :>25ơ .@ ơ /@ ơ%ơơ !ơ>25ơ )ơ µ0ơ ō 2> :ơ ơ +-./@ ơ Ŏ :2sơ 8 +-./@ ľ ơơơ ,-./@ ơƀ-
B7ơ
'H# ơ '"#%@ "%@ ? "#@ ơ ơ ơ 4 ơ ơ *"@ ơ #%@ ơF ơ ơ ơ ơơơơ ơ ơ #@ ơ "&@ Uơ "#@ qơ#( @Yơ "(@ 9wơ A ơ ơ ơơơm
'=ę A ơ ơ ơ
Q È ' w
j7p
'7# [ ơ ơ ơơơ ơ 4 ơ ơ ơ ơ ơ
ơ ơ qơ'ơ'Ī# # Uơ ơ ơ2@3 @ @5 @ @6@
2@
3 2@
6 2@
@2@
XZ
YZ
5 3@
6 7@
235
A ơ 376@2@ 'w J 4ơơ ơ ơơ
=) º» ZÞ
X
@ Ɛ
ơ ơ ÓÞ ? ơơ ơ ơ ơ4ơ 'wơ A ơ ö < # Ě Jơ ' 4ơ ơ ơ ơ ơ '"#%@ Ûm ơ "-@ ơ ơ ơ
ơ #%@ ơF# U ơm ơ $-@ ơ %-@ ơơ %"@ ơ "#@ ơơÜơ ơ Ėơ "-@ @9ơ)-@ @ơ$-@ @Gơ&- @'ơ ơ %- @ơũ ơ ÜQě 9# [ơ ơ ơ ơðơ ơ2:@ ơơ4ơơðĎ ơ ơ ơ4 ơ ơ ơ ơ ơ ơơ'ơơ ơ 9ơ ơ ơ ơ ơ 4 ìơ ơ 'ơơ 4 ìơ ơ '#ơÛ ơ 2 Y( Jơ Þ 4ơ ơ ïơ ơ ơ ơ #ơ A ơ ơ ơ ơ ơ ơơ ơ ơ mơ ơ ơ ơ ơ ÖÞ ơ Þ ơ ơ ÓÞ@ÖÞ ơ ÓÞ ÖÞ
4 ơ4ơ H# q wơ Jơ .@4ơơYo ơ mơ wơ A ơơï4ơơ ơ Þ ơ4ơơ ơ ơ .@ ơ ơ ơ ơ ơơ Þ ơ ơ 4ơ ơ ơơ ơ ơ ơ .@ Yơ
'1?-)BAE+Q $)H.+<)H1*)9Q 'A*1+HIQ # # ! # #
Z
# , &! , !) , JÞ !)(! %, X{ +ªơ ' ²ơ 1\ ơơơơ VơơCơ"ơ ¡'ơơ9ơơơ '¢ơơ 919\ ơ ơ9B CN,-.\ Cơ ơơơ ơ 1\ ơ ơ :19\ ơơÁ" ơ ơơơ ơ §ơ Þơ ơ 919\ <ơơơ »ơ ơơơơ ơ B\C\ ơ ơ ơ ơ "ơơ ı
¡'9 cC¢¡G Ó=7X¢ ¡7G7 Ó7=77X¢ơ ơơ 9BCN\.Z cơ ơơơ"ơ 'Ğơ Ì
+Õơ cG= ²ơ ơơ ơ8 ơxÃơ"ơơ Vơơ ơ íơơĜơ ²ơơ8 ơ Á Vơ ]ơ V íơơ ơ xýơ ơ ơơ ' 3]ơ" ơxħơ ơ ơ ơ ơ8 ơ
''7 ''ơ3 ]ơ §ơ
]ơ
ơơơ ơ
XÈ ]ơ
XX7 ''ơ3 ]ơ #Q
s6O
]ơ
X 3]ơ
cG={ơ
9{ +ªơ C9G ¬
ơ ơơơơơ ơơ 1Þ ơơơ"ơ8Vơơơ ơ ơ ơ ơ 1Þ ơ ơ ơ ơ ơ Þ ơ ơ Þ %' ơ ơ ơ VV
ơ õ ơ W XơÞ cơ W c
7 7
ơơơơơ ơ"ơ Þ ơ 'ơ W XơÞ ơ W
7 õ7ĝ c
Þ ơơơ ơ "ơ ơ ơ Ļ»Äơſ ơ ơ
7
Þ L=Þ V YMÞ X<Þ Þ KÞ
c'ơ
k>bn2Þ
cz
ơơ&ơơ ơ&"ơ"ơ{
"ơơơ "& ƛ & Íơ ơ c)
N
*
c2
N
3
ơ"ơ& ơ&ơ 3
cz u
N
cz
{
q¹XCL» 7» "ơ ơ ơ " ơ ơ V(ơ Uơ &ơ &ơ {
+
"ơ z@»666»»z» " ơ& &ơơơ "& & (ơơ & ơ& ơơzD»666»»z» &ơ cz
&"ğ ơ & ơ z@zC»
ơ ơ &"(ơ ơ & ơ z@zC»
ơ+
3 &"ơ
*
{
"ơ & ơ "& ơơÁ""ơ & ơ"ơ*
N»"{ơ U"" & ơơ & {
z@zL»
ơ ,
3
{6
&"ơ (ơA&ơ"ơ"ơ"ơƃơ ơơ ơ czN*
/7
N7 3{6
{6 N ,
/7
{J *
+{6
{8
NO
*
37
{J *
{J 3
-7N *
37
. s}»s£Y» ơ
x
Z
8
3A88
+P ơ Vơ ơ ""ơ &ơơ &ơ & ơ &(ơ àơ ơ ơ ""ơ ơ &ơ &ơ ơƗ&ơơơ&ơ ơ ơ &ơ"
B
3 07 17 3;3"
àơơơơ ""ơơ&ơ"ơ "
ơ ơ&ơơ ơ &ơ" 7"
47 27N
6*; ŀ ơơ &ơ "ơ B
3;3"
6*;
O
3A88
; +P 3">B*
cV» ©»» c» <
i»«^» c»L»N
©»» c» 3
h» !»<
¡»«^» cR»N
ª»» c» 6
i»»<
¡»«:»
U
U
ơơ
K
)
Þ
)
K
)
N
v-(=#$
q
N
*<< K
icME»N
w *"3""
$7x *%=&'
%7N
6 6 8K
%7 *<>
icRE» y *"""
#7 6 83
icVi»
8
<
Aơơ ơ B\ C\ơ4ơ 4?ơ4ơ ơ » ơơ4ơơơơ ơ 0& ơ
» ơ ơ ơ 0#\ ơơơ ơ ơ 0$\ ° ơ «ơ
r
<
*<<3
7 \,<<
6&&& ,<<
\669
,
W
,"<3<
P
3"
+P ʼnơ
\ÖÞ \ \S \ ơ
VZ ơ ơ?ơ ơ \ÖÞ \ \U&'Z ° ơơ ơò?ơơơ4ơơo¹ ơ ¹ơ? ơơơ? ¹ ò P
3"
"
\ÖÞ \[ \P \R
Þ
ơơ ơ ơ Q
?
P
,"<3<
P
3"
*
+P ļ ơơ ơ ơơ Ĵơơơ ơơ ơ ơơ Ġ ơơơ
,7 ;
\+7 *7 *
K» ;
Q
*
A
T
W
W
+Pơeeơ +ơ ơ
fZ\^g
P
,!
ơfV^Zh
«ơ » ơ fY^Zi
Q
,,
U ơ U[R
?[Y »R
fV^Zg
R
@fY^Zg
P
pp.Þ
B
J
B
<;99
3
"* 0K\ CP
T*
Z[» » ;"
C
6
9\ *y»U ơ 90QN\P
*<
;0!9\ *?
ơ 90#9\P
,?
Ńơ ơ4ơơơ 6.
B C D\ ơ ?ơ ơ ơ TBCDU\ 0K\ ơ ơyơơ TBCDUP0K\ ab ơ ?ơyơ Q
, 3 6
ơ ơ ơ +P Ay«ơ
&'Z
WZ
Z
96
;
'( + ·ơ ơ ơ
!Z
tq?ÛÞ
Þ
**
uq?Þ
totÞ
+İơ 7 U l ơơơi ơ ơ ơ ơ i ơ i ơ
Aj
F7»
Hơ ơơ ơi şƅ ėơ ƊæơƔæ k
l 2
N
j * 2
k
kmO
j2
~j ơ ơ k
ơ ơ ƆƂơ }2
j
j2Aj
W» S
(ơ
jHI ŗ ơ ơ ½ N
fj * g2 :j2Aj
Hơ*}
5ơơi
$ 5ơ F» FG
7(ơ [ơ k
N
l
N
:
j
N
7ơ ơơơi (ơ ơơ ơ ơơ j
ơ 7(ơ +ơ *;"<
S
L M
FG
*
ơ ơ ơ ơ {
N
C'ơơ ơ jo
S
y ź a©HŅ(ơ Őơ ơ G©ơ S
ơơ jp
6 (ơ U ơj0
N
ơ ơ * $$ơ ơ ơ ơ ơơ C'(ơ ·ơ $ơ ơ ơ ơ
ơ (ơ5ơ ơ ơ ơ Cơ*
"":
N
*;"<
c
c
Z
mÞ
'(
2
343ÞÞ¢N?ÆcÞ
¢N?ÆcÞ
¢N??gÞ 544ÞÞ ¢[ÆÆqÞ
* 6
+ FF N» U ơ ơ lơ
ơ ơ ơ i ơ ơ F» ơ ơơ lơ ơ lơ ơ ơ i (ơ +ơ ơ lơ Z ơơơ (ơ [ơ ơ lơ ơ ơ ơ ơ (ơÝ Ĉơ ơ ŧ ơ ơ ơơ ½lơơ ơ ơ ơ ơ ơ P» F» ơ Z (ơ Ý ơ ơ ơ (((ơ Ż©aơņơ Þ NAA rel="nofollow">FUO»m» ''(ơ
j
k
jk
jk
jsk
S
jsk k
*
2
k
2 k
jtk
2
jsk
N
{
jT
k
jN
k{ HI
{
k
{
k
k{{
2
HI
k
{
j
*: ơi ơơ lơ $ ( XC( +ơ C [ ơ ơ ŵơơ/Iơ /sơ ơ/ơ ơ /ơ /Sơ ơ/(ơØơ ơ ơIơ /®aơ 3 /Iơ /ơ ơ/ơ aơ/ z/saơ Þ /Saơ 3 /sơ /Sơ ơ/ơ aơ aơ sS /(ơ ơ I ® ơ sSơ ơ/(ơ ơ ơ ơ ơ ơ ơ /ơI5ơơ®ơ ơ ơ ơ ơ ơ ơ /(ơ ơ/ơ sSơ ơ/ơ IaÞ
X]2
N
`2
N
;5
N
N
A
ffÞ
S
6B
+Pơ ' ơ uơ gơ ơ
ơơ QS(ơ ơ Q!ơ uQơ!ơ ut(ơ Jơ \Qtơ!ơ " $ ŊœQuơ!ơ \Rơ!ơ =6ơ ơ\RQ/ơ!ơ\Sơ!ơ '6Îơ +?ơ\uQtơ! \/!ơ Ò=Wơ!" ÏÞ ??ơơ\tQơ!ơ Ò=Wơ36 3=6 3$'=Wơ HÞ"$ "$ ơơQơ!ơQt( åơ ŌQtuơ ơ8 V?ơơ\uQtơ!ơ '=W 3!"$GWơ ơ "$!ơ 'W(
b
+Pơ JơFI!ơ "$ /Sơ$ $ ơ SF!ơ ơ ĹFơ! $"$ ơ \S/Fơ! ơ\Ľ/F!ơ $ róơơ ơ ?ơ? ơơ Į/FSơơ ơƇbơ!ơ$Þ "$ !ơ 5» $$" $ ř ơơVơg ơơơ $$"$Þ $Þ "$!ơ$_ơ"
@
åŰơ ơ gơ ? Ū ơ ơ "$$ " $!ơ (ơ U ơ $ > $ ơ ơ ?ơ ơ ơšơ $ơ ơ $"$$" $!ơ ơơ 8 ??óơ "$$$ \" $!ơ (ơ ơ "$!ơ $ \ $ Aơ ơ ơ g ơ Ĩ=Î
@ F @C @F @C
IF!ơ "$!ơ ` `
+Pơ őơơL 7W 3¸ơ!ơ$¸ơơWơ 1 ¸ơ1 7W ơơơ ơ ơ ( U ơëơ!ơ$3 !»
QL ď f»! QL 7 3L ë 3 f»!
`g»
`g»
$
$ !ơ $ u`»L Ň 3'
1Ô# +Pơ ơ ơ ơơ ơ ơ ơơ 0/\ Cơ 1Ïơ/\ 9#ơơ ¾
ơ /\ MQ ĀC _ơ1ơ \19Æơ/\ 1CEơ ± ơơ ơ ơ 0\/\ E # +Pơ K 5
D\ B\ C\ \ \ ³´µ» ơ ¢ ¤» ¢¤» ơ ơ ¢»/\ ¤»çơơ » Z #ơ ơ ¤_
CB\ B\
DC\ C\
EF\ D\
± ơ CB\/\DLC\/\ELD\ 5ơ B\C\D\Ţ ơơ ơ¾ ơ B\RB\R"B\R$B\
ơ CDELB$\/\R&\/\ R$"\/\ 1# +ªơ ä ơ 1Ïơ6ơ ơ¾ ơçơ6ơ ơơ Tơ ơ ƒ
K T ơơL6¤ơơ. ơL6¤ơ E
+Pơ śơT
ơ
<5\ 05\
NQ 6ơ ± ơLŸxxx¤ơ/\ # =0<2@\ >024?\
=2<4?\ <6\ =0<4?\ <8\ =024?\ 26\ =024?\ 48\ <5\ <7\ <8\ /\1ơ \ \ 05 26 48\
$
9
<6\ Eơ
B
9)
+Õơ }9== [ơ w ÝÞ %\ ơơ BO\ ơMơ ơ CO\ ơ ơ.ơ Dơ ơ ơ ơ @)ơ rơ ƓMơ ơ .ơ Dơ ơ ơ ơ ơ ơ ơ ơ ¿ ơ CO\ + ơ ơ.ơ Dơ ơ ơ ơ ơ ơơ %\ơ.ơ BOMV\ ¶
+ơ ơ CO\ơ ơ ơ ơ <;ơ ơ .ơ Dơ ơ ơ ơ ơ <<ơơ COA\ ơơ.ơDơơ ơ ơ ơ<ơơ 9<ơơ.ơ BO#V\ ơ £ÁÞ ¤Â GÞ Þ V¡Ã!WÚÞ
ƝƁ Mơ ơơBO/\%BOM\Çfċaơ K ơ B \ |Ĕ
%\ ơB#\ @;ơ ơơBJ\ H99G=)
+įơ @} ´ơ ơơ. ơ ơ ơ ơ .ơ GơơD Ö z;ơ z<;ơG;ơ z;ơ9} ;ơ ) ) ) ơ üơ z<9;ơ<ī)ơ Kơ kơơ@}ơ Eơ ßDơkơơơ ơ ơơ ơ . ơ .ơ H;ơ ơơ Eơ Œơơ ơDơ kơ ơ ơ ơ
ơ .ơ H)ơ rơ ơ ơ Mơ kơơ ơ ơ ơ ơ ơ
. ơ.ơH)ơ ơơƙ@;ơ;ơ) ) ) ;ơơ<|ƚơ ơơơơ Mơ Ó Þ×Þ ơơ ÓÞ ×Þ ơ ÓÞ & ÕÞ ơ . ơ.ơ H)ơ
})
+Öơ H| [ơ .ơ Mơ ơ ơ ơ ơ ơ ơ ơ ơ w M ơ v ÝÞ Ô#ơ Ŝơ Ũơ ơ ơ .ơ Dơ ơ ơ kơºơ ơ Ź O ƍơ: ) ơơ.ơơ *) \ ơ ơ.ơ. ơơD Eơ âơơ .ơ :>2ơ ơ ŴEơ ¶ơ>ơ ơ ơ ÷ Døơ .ơ ơ2ơ ơ ơ ùú .Eơ ơ Žơ Dơ Dơ>ơ ơ2ơ ơ ¿ ơ ơ .ơ ơ ô .Eơ r ơ : M ơ> M ơ2ơ ơ ơ ơ ơ ơ ơ Dơ ơ ơ ơ ơ ơ Dơ : ą ơơ ơ .ơ ơ .ơ ơ Dơ ơ ơ w & Hơ E O ¶ơ ơ¿ ơơơ W\ +) \ H|)
|Hơ
'1@-)CAE+Q $)H/+;)H1*)8Q 'A*1+HJQ # # " # Q
" , &! , !) , , &)$*, , ) , ,
,
%&$)&! %, &!, ! &%& &%, O: 3JPSANZ 399Z iÞ MRAPQEKJP
:KZ ?=H?RH=QKNPZ =NAZ =HHKSA@
@E ơ fp`ơfb;ơ) ) ) ơ`ơfơơ w ơ ơ ơ žơ ơ ơ $ ´Tơơ ơ ơGZ VZ `ơ ơÀơw ơ6 p`ơ6 b`ơ) ) ) ơ;ơŝ`ơ ơơ ơT ×ơ GZ
ơ ơ ơ$ ơ
"=Q
_gSh
i6o
-Z % ¯»
ºÞ
FÞ
-Þ
) ßơ fħMơ = 44 Z fơơ ơ ;ơ Tơ ơ 6 Č fp6 č fb») ) ) ơ6 3 fE ơ ơÀ ơơơ ơơơ Mơ ơơơ Ŧ ơơ ơơ ơ@E 9ġ ơ Zp;ơ Rpơơơơơơơ ZRơơơ ơ ơ ZR2 ;ơơ ³2ơ = #$ ơ ơ ³Zpơ2 Rpơ8 ³ZR2)ơ Þ ơ À ơ ơ ơơ ơ ĭZÑRÑ2ơ ơơ ơơơ2 Z ơ ơ2 R ơơ ơ Zbơ ơ Rb;ơ T ) ´Tơơ Zb Rbơ ZR)
i
ơ Þ ơơ ơ ơTơ`ơ){`ơ Þ ƘN¥ơ `ơ 4 Z«;» A ơ ơ ơ *E Þ ±» Þ ơ !*6E$ *7E 6E$ w ơ ơ q w /Þ
j0
A ơơ ơTơơ6ơ ơơ6 ơ T ơơ ơơTơ
\» ²®»
Y=ơ
'1?-)DAE+Q %)H0+<)H1*)8Q 'A*1+HKQ # # # #
Z
" , '! , !) , TÞ !)'! %, @ĕ Ş ơ ơơ ơơơ ơ ơ$ ơơ &/Eơ ơ ơ ơê ơơ ơơ
8E
ơ ơ CE&/E *Z l3» ơ ơ 2 ơ 8 ơ ;E;EEE;8E E (Z ;/EE 2E31ơ ơơ ơ ơ8 ơơơ E (Z "D 4E&/;/EEE2 1^ơ2»
ơ
ơơ $ơ$EE2E¦1^ơI?\ ơ28E31ơ8 ơ#ơrơơ $ơ $ ơ ơ m ơ ơ ơ 8 ơ >>EE>8E ơ @@EEA8E ơ Êơ ơ ơơơ ơ ÂÂơ ÎÞơơơ ơ ĥ# ĉ #/ A/EơŶơơ &/#/ &/@/E E 335 ?:E 9%E #ơ Kơơ
U+
Ŕ ơơơ ơơ = &= &E = &8E31ơ!ơ *=+=Eơ¼ $ ơ *=E ơ +=E ơ ơ è ơ ơ ơ*=E ơ+=E£ơN#ơ Ŗơ *&/+&/E!ơ ¦@ơ Ů!ơ Nơ¥ơ#jjơhơ $ ơ ơ *&/E!ơ Nơ ơ +&/E!ơ 3Nơ ơ *&/E!ơ ¦1ơ ơ +&/E!ơ 1#ơ ơ ơ ơ ơ -=E!ơ *=E_ơéL6^ơơ -&/E!ơ E ơ $ơ DZ!ơ 1¥ơj##ơhjơ +ơ -= +Z mÅ LL6^^ LéL6^^^ơ 1 hơ Ƅơ ơ8 ơ -=E!ơ E ơ hơ ơ#ơơ $ơơ -=E ơ ơơ ƕơ $ $#ơ ơ *=E!ơ +=E ơơ
ơ» ơ$ơ$ ơơ6#ơØ ơ ơ ơơơ ơ ơơ ơ ơơ ơ ơ6ơ ơ ơơơ = &0= &EE= &8 @ơ ơ jơ #5*$!.-$# 5 5 ơŤơơ ơơơ jơơ ơơ ơ -=Eơ ơơ )ơ âơ *&/E!ơ F»F» )Z D,Z 2E ơ *&/E!ơ (F» 2E_ơN×ơ ]» hjơĂ +&/E!ơ 3@ơ1ơ (Z D+Z 2E ơ +&/E!ơ Năơ 2E_ơ@ơ]» 5+Z h#ơ ơ ơ*= 1^ơ! !ơ °# K ? ŭÅƋāơ £ƠĊ Ŀ (),*=E £ơm 2h 2E £ơ
ơ*= (),+=E hơ ơ ơhơ ơơ ơ ơ*=E (),+=E 2E °$ *=E31ơ!ơ '= &= &E = &3E ơ += _ 1!ơ ơ '= &0= &E = &3E ơ¼ơ 'E'EI ;,Þ ơơơ *=+=E_ơ*= += Nơ!ơ .'$= &E=B &E = &31E rơ ơ è ơ ơ ơ =8Eơ''E!ơ N#ơ ơ ơ *= @ơ!ơ +=E_ơF8» K ¼ $ơơơ !=E_ơ@ơ!ơ += @Ćơ ơ Ê
Y7ơ
9E ơ ơ ơ ơ ,ơ ơ ,ơ ,ơ ơ gơ !' A ơ ơ ơ ,ơ ' Þ À¨ ' ơ -ơ ơ ,ơ ơ,ơơ !' ,ơ '' ơ ' ơ '' ơ' ªÍÞ ' {» ' ' ' c !' ¦» ơ !' Ô)Þ ơ ¦» \ÔÞ {y~»G{» \c 1=WËơ ·ơơ !' ' ¦» !'' Ô*Þ gơ ơ ! ' !' G¦0» ° ơ !' ¦» ơ ñ !' ¦» \ÔÞ {1» ơ $ ơ ơ ' !' ơ ' ơ $$"-ơ K $ '!'ơ ' ơ -ơ ơ ' 'ơ ' ' ' ' ơ ' ' ơ ' $' ơ ơ ' ' ķŷơ ' ' ơ ' ' ơ &' ' ' ơ' ơơơơ ' K ơ' ii» ' ơ ' ' ơ ñ-ơ
a
#5 )$ .+$#5 ,ơ " ' gơ ,ơ ơ ơ ' ơ ,ơ ơ gơ
' ơ
,ơ ,ơ ơ ÉÞ' ơ ,ơ ' ơ ,ơ ,ơ ơ ơ ơ ơ ,ơ ơ gơ !'
ơ ơ ' ơ ' ơ ,ơ ơ #' ơ ơ ' ' ơ ,ơ ' ' ơ ơ ' ' ơ ơ ơ ' ơ ' ji» ' ŏơ ' <=W 3 ' ' ' ' ơ ơ ơ
'ơ ' ơ -ơ K ơ ' ki ' ' ii» ' $ơơ,,ơ ! ' ơơ, ơ Eơ ơơ !' ' !' ! ' !'' ơ $ơ ơ !'' ơ ơ ,ơ -ơ K"$$ơ'!'' ơ ơ ,,ơ -ơ
ơ ơ ,ơ ÊÞ ' ơ '' ơ ' ơ ' ơ ' ơơ $,ơ ' ,ơ ' ,ơ ' rơ ä ûơ ơ ơ # ' ơ ơ ƌơ ! ' ơ ơ ' ' ơ ' $$ -ơ Ř"ơ,ơ ơ,ơ ' ' ' ơơ ' ' ' ' K $ơ ' ' ơ $ ơ ,ơ ' ơ ơ
ơơ ' ơ ū ơ ơ ơ ' K ơ ' ơơ '' ơ, ơơơ ơ ' Ŭơ ' ơ' '
qơ
ơ ÉÞ \ ơ ơ % ơ ơ '1\2 \4\ ơ ơ ơ ơ \ ơ ºơ ơ ơ G#\ơ ơ #\ ãơ ơ 41#\ ơ 42#\ơơơ H#\ ơơ 41#\ 42#\ ơ % ơ ơ '412\ \ (41#2#\ ơ ơ %ơ ơ ơ ơ 4\ ơ ơ 41\ ơơ%ơơ 1 \1#\3 \3#\ ơơơ ơ 1X\1Y\3Z\2[\ n#ơ 1\ĸ ơ 4\ %ơ #ơ ơ ơ ųơ nơ ơ ơ ơ ơ 4\ Ēơ ãơ ơ %ơ ơ ơ ơ 11 2 2\ ơ ơ % ơ ơ '4\13\ ơ ơ ơ ÉÞ ơ ơ %ơ ơ ÉÞ \ ơ ơ ơ 1\ \2\\\ ơ ºơ %ơ ơ G#\ ơ ơ ơ 1 2 \ 14\ ơ 24\ ơ ơ ơ ơ ơ ơ '124\ ơ ơ ơ ơ 14\ ơ 34\ ơ 1Y\ ơ 2[\ n#ơ ơ #ơ #ơ >20Z >51Z \ 65 Q\ Þ ơ 1\ ơ ơ 2\ ơ ơơ E(Þ ơơơ 62 Z 72Z 75Z RÞ ơĦ ơơ %ơơ 72Z 75Z % ơơ 1#4\ ơ 2#4\ n#ơơ 1#2#\QQ `È§Þ ËĽÑÌ´Ä¿wÞ
e'
Śơ ơ ơ
, ơ ơ nơ #ơ [ơ ơ ơ ơ , ơ ơ ű ơ
,:QQ,>Q
ơ 5
<
,:QQ,>QQ,>QQ,>Q !,:QQ,> QQ!,>QQ,> Q : >> >Q :>Q
ơ njơ[%ơơ ơ8ơơơơ8ơ
,,57,5 QQ575 QQ5 Q
ơ
,,5,5Q 55QQ5Q
µ ơ ,Qơê ơ ơ
,5Q@Q ÆơQ,5 QQ,5QQ,5Q
ơ
,5QQ ,5QQ,5 ,5 Q
ơ Ɩơ ,Qơ ơ % ơ ơơ ơ %nơ ơ 2>QQ{»Q > 6FQ ơ {»Q,6Q ơ FQ ơ ơ %%ơ ēơ ơ ơ 8 %ơ {»Qx{»Q: 7FQQ{»Q> 7FQ 6(FQQ:QÞ > Q ơ % ơ Q{" SÌ B» Þ :QQ>FQ :QQ>Q ¬%ơ ť ơ ơ ơơ ÌÞ @ ơ {»QQ ơ ,Q> Q QQ> Qơ % #ơ ¬ nơ ơ ơ ơơ ơ 8#ơ
Q
®$Þ +] ?IVG] /I] %
i
M] ^i «ØÞÞ i,] 72] M-] EG/8I] >bc?[HDBi /?]/?G I/?*] \cBDc i Y] GMBQ9HI] CJMR2M! V] -U"] zÞ 0
0 _i (CG] 22]
²Þ
H
¬Þ M+RK] ]E -U#]
0 _Ni 0 9OSi 0 #`Ni
PÞ V-/-
/:E2/I] N.N] !_L);2iÞ ~Þ 8Oi k L
;Z ,» 0 ÐĄơi iZFZ REECK] )\ci IC; ] FZ G #Þ DM$] M-M] ÇyÞ e» )\ci ³Þ FZ Z Þ M-? -» Z hÞ NÐ)ơ /?%] NBÞ ; $Ti ?] B Nơ ~Þ xPÞ V] -U] %.i 9\ . /?# FZ Z Z GAB^M+E5P8Qi PÙÞ &-i Nơ Ò»Þ M-RJ] Q_xP¶¼«¹P e» ( ./I] :
M] µÞ
y
Z
"gI4Ji CD&PÞ ~Þ
/M/C?3=J] ·Þ 0
Z ?¥Þ
*
'Ti KYaUfK[Gi
~Þ 8Qi
0
» 0 i ]ci &} 0 -*Z
¨/» V-/-] /I] ] C?NG\
i /I] 0U/J/2'] Z] 62] ECI/P/U#] /?M*"GJ] ;Z
» M-SI .» ~Þ " i M] M-?] )C22CVL] M-M] !z /J] /U/K/2] Z] i |Þ Z "i i PÞ ®Þ "i ,i /?22[] JTEECK#] v» Z 1 i Þ M+@] -U] $Vi {PÞ Z "i e» /&] &} Z (Wi x V-/-] /K] ^ C@MG/M/C?] I/?] + 0 Av]^B%Þ 5>J » 0 r}Þ CG] / 0 "1 i /?#] i e» - ?
XM] d ?CM/ ] M-M] M-] K[dCGDci
Z i :
$| 0 "<i V] -U#] #{
}Þ i
UMAWM/U fi J/? ] '~ 0 *i ¾¦Þ i 07i #i ! i
d
V] C?2Z] eh@Bi MC] -1] M-] IJ] "z
!( $ $ ( $'( ( " " ( & ( ( % (#( )' -/ /
-/ /
/
!% (( #!"!( " (&# ( !% !(("( !% (!"( $(
( "( #"( ( #!"!( " ( &( "( "" !( ( & / ( / * / ( ( !( "( "( "( !% !(("( !% ( !"(
( "(" (! "(#!"!( % "( &# ( !% !(( !% (!"( (! ( "( "(#' !(%( &# ( !% !( (!"!( (("(#!"(&# ( !% !( (#!"( !( ( ( # " !( ( %(
(/ &/(/ '(/(+$/ ('/ $/ / / / / % / /
' / /
)/ / + //
/ / $/
,//
(/ / /& /#)!&/ (/'/(/"#")!/* )/$/ ./ # % //
' / /
)/ /
+ /
-//
"'/ ) ('/ (/ ')"/$/(/ &'(/( %$'(*/ #(&'/ #/ #'/ ( (/(/ ')"/'/ / / '/$)#(/$#/#(&/(+/ +/ $#/'/(
% /!/
' / /
)/ !/
+ /"/
-//
/ $/ (/$ $+#/'/ /%$''/#)"&/$/ $# '/$/ /$#*,/%$-$# % / !/
' / !/
)/!/
+ /!/
-//
(/'/(/ &'(/%$'(*/#(&/ ( ' ('.#/ / % /!/
' / !/
)//
+ / /
-//
/
< 6ę J"ę&ęęPJę"ęÛŴ JJ ęęPę>% Ŵ bęJęę^ę ęęêŴ ęęĆ&%Ŵ fi
>i
,O
9
Ci
Ri
=i
?i
!i
Cę @ i
dä
2Yę
/ bęJęę^ę ę ę pę ľ ľľľ eiľ W eiľ
îę . 2ęľ
2ęľ
ę eiľ W eiľ ę eiľ 2ęľ i
>i Sľ
Cę ,Oę
Cę @ i ľ
Cę Bi
Cę 2Yę
/ ľ.% ę 6
9$ ęę(ęęęę&J"ęJę2! .ę(r ęęrr^ęr"ę
>i ľ Kę.ÜÝę Cę ,Oę Ćľ / KęSÀÁŴ ÚÕÞę ¬ä°ę ąľ ÏÖ ´ę SŴßę //C Bie56i 9 !ę.ęę 9 Cę Y±ę ęę Ò µę śŴ àę Ù Ç $ éę^ę ę
ęľ ę
8x .%&"#/i
ę ę 9ę ľ ľ .%iľ ęj . Ôęľ ę 2iľ
ľ ę
õþľ
>i
ę
{
^_ę
,Oę
ę
{ä
8MM× .%&ZZai Cę
@ i
ę
{
8 MMx .%&ZZ/i ęCę
Bi
ę
{
8MM .%&""1i Cę
2Yę ę ęę
$ ęę[i ę\i ęJęę ę(ęęęęęPJę e% *ęę ęľ bęJęę^ęę Diľ vä \i [i
>i Cę ,O
@ i
Bi
2Yęę ę^
ę
ęľ:ę
ęę
bęęę đę&ę-
ę )ľ -
ę )ľ -
ę ľ -
ęę + ę(ę;p
- U
ę --ę
Zę
ľ ľ
P ehv
Aę&ę [ľ Pi ę Aľ ę+ę"Aę öÿ ęę+ęęľľPiľAľ
ę
ľ ęę+ęęę @ ęęþę+ęę ľľ f ľ
"Yľ
ľ - ę
@ ęę+ęGęęęę
"
- ę
< }ęľ ęę
Yęęę < ęę7zä Y
ę Ø ę
ę < ę
ęÈä
-
ęfę
-
ę fę
@ ęę ę&ęľ ę + ę(ę Ŵ ę < q+ęędi 7i ę ęęġdc ľ !ęi ęę+ęędi g $ di di -I
» &Q %ęę% (ęę ę ę%e ęę k!ę &ę%ękę%č"ęę%ę U &Q ęęę%ę%ę ęęęęęę%!ę&ęęęęęę
kęę%ęę%e p ¼ { ęę&"ę¥¦ęi"$
q+ęę G[iľ G \iľ ńŴľ ŋŴľ Gci ľ
lę ę Ŵľ ÒŴľ %Ŵľ Ŵľ &Ŵ ľ ę ľ
ĝ \ľ i ęę+
$ q+ęęěľ ęę+ę"ę ęIę ľ Ŵľ -ęľęľ ľ ľ ľ ľľ 6ęïòù¢ęęIę %ę ę7ę @ ęęęęę
ę
- Zę:ę Tv / Wxę / <ę @ ęę' 'ę&ę:ęę + ę(ę
<
- Zę?HFDRSęęęs"ę%ęęę "ę?DLęHRę ęFSę'%ęęęę ¥h ęę'ę*ęę'"ę *' ę(ę (ę 'ę %ęęę -ę2*'ę sLę 'ę * ¡HFDęę-<ę @ ęę'ę*ęęs"
-- ZęFę ę ę %'%ę &ę ' ę -
$ę Tę )ęę 'ę % ę ę ę %'%ę ę %ę&ę(ę&ęęs% ę-
7ę êęęę'"ęę)pę
-ę Zęę .ęę+ę'ę'ę%ęęwęľÄäľ ®ä ľ RŴ @ ęę+ę* -/ę
Ąľ
-< åŴ ę PLę IęúĈĉľ ę ę 'ę *ę ¡ìíę Uę ę *&"ę i"'Lę *ę DRęęęHFL Ię#ľ ľ ę ę õŴľ ľ Kęi IęĂèê ľ
éľ
çľ
-< bęęę' %ę*ęę'ę'ę*ęęP Ðę ľ ęľ- -/ę ę
ľ
Ñę ,ľ ę ľ - «ę
- < 6'ę'ę*'ęę*ęÛę ę&ę ęLę &ę&ę9Kę ęę&ęę @+ę(' ð' ę Kę-Kę³ęę ęęeęĖē!ęęę' Đę*ę'ę'Lę%"ęę*ęę ę +ó"ęęJę&ę'ęę 6ę'ęęę('ę&ę%ę ęę Tę 'ę*ęęċ('ę&ę -/ Zę Ęľ ËŴ )ęñę &ę'ę'ę %ę ę :Wę :) : w½ )w
-9 6'ę'ę ęõ&ę"'ę+ę*ę ęė%ę ę"'ę+ę*ę
ę
Tv
)ęľ -ę ę )Wę
W m -ę
ľ
:ę ľ -ę @ ę ę +ę *
ę ę"'ę @ ęęę*ęę
h&ę(ęę*ę"ęG!ę.ę(ęęP
bäľ Ïä
Tv
6ę ªfę ęę(ę%%ęę!ę ęUę%ęęę&%ę%ęę " $ +!ę(ę%%ę ęę (ęę "ęę*ęęęę*ęę&ęę 6ęU %ęęę%%ę ęľ ę ę ęęęę ęę 5!ęęę*ę("ę ę(ę 9ę ę ( ęęę ę *ęę ęęę *ę ("ę /ę(ę /ęę ( ęę ę ę*ęę U*ęUę(ęę+ęu "ęę(ęę+ę&u "ęęę (ę%%ę ęę (ę Kę&ę(ęę% ęUę+ę zę
6ę*&"ę%"ęę* ę(ęę%ę*ęPęę ÷ęę Tę %ę ę*ęęPęYęę*ę"ę/ęôę Zęęę*ęę%"ęęę%8 ę @ ę +ę*ę$
æľ
äľ
I
îľ
ñľ
ìľ
đľ òľ
ôľ
Tęę ľ ęę+ę"!ę ęOiPi ę+ęęę&ęOiiPi Sv $ę @ ęęę+ę* ę ę
ę OXi
,I
ę iPYi
bęęę"ę+ę"ęľ *ę&%ę)Eę i
ęę +ę(ę ľ i z
ę
Tęęę&ęęęęP ę
ę -ę ę ę ę V[ -ę 0 V[ ę 0 V[ ę ęę ęę @ ęę+ęęęę<ę $ Tęę !ę1Aęę .ęęęę3ę ę1.ęAę
0ęę1Ąęę;Aę
Eęę1Ääę ę ę
@ ęę+ęę oęę1.o7ę
ę
Óęę1Éęę $
3<.$A?C*Y $F0*9$F3'$6Y ?'3*FIY & # # & % & & 8 "Ŵ $ & #& $#"& < 5Nę G5 O3ę
ęęęęę37
< 5Nę GO aę8ę ęię Tv 2 8ę < 7 5Nę @ aę ę i ę i7 7 ęľ i
ę Tv $ 5Nę A
¶ )2)ęº ę 6ęęę "ęęę) ę("ęę 7ę h3ę A ę 5Nę @ O3ę)8ę 0ľ Eę Tv $
5ČNę A
}ęOięę"ęę>¢Ŵ 6 ęDRę ęęøDRFę
VOi Oi ę l 8ę iOi7ę ęËl VOi ľ DI
Tv
CI
Rv
«pä
; 5Nę GO a
ę 2ęi2ęi ę ę
Liię
ę
ę iii 2ęi 2ęi¯ 2ęi
2ęi
ę
ę ę ę iii Lii ę Lii ę Liię Lii
ę
ę
2ęi2ęi
;ę
ę
Lii
ę
ę T (%
Liię
Tv
ę
Lii Ê ę
Tv
9 5=ę G5 T3ę0ęi¯äęę34ęP!ę4ę4ę "4ę4ęę !ę!ę!ę!ę !ę$ę h3ęG{ę ę A i 4ę4ę @ę GOę ę G!ę 4ę 4ę4&ę "4ę ę ę ę ę ę h&+!ę ęęę ę&4ę !ę 4ęPę4ę ę&4ę4&ę[¨ę 6ę4ęęęG5ę&ę i.0ę 4Ēę4 4ęę99 ę 5Nę A
T3
c
ą Ćę 0d c 0dę i
ľ
ľ
c 0dę i f
c 0d ę
c
ć Ĉę 0d » 0MM
c 0 d ę
3ęę43ęę 6ę4ę4ęę
ę
ę
$
FI
?@
A 0¼ Ev ę
5=ę GO
[i i\i ľ $ę 6ę(ęęę4ęę !ę; 5=ę $
ę >ľ
ęľ
ęľ
ę Hv ę >ľ ę ľ ę Hv ęG ę;$
$ 5=ę $ Y(ę334ę&ę44ę ľ !ę 1ę ľ ę ęĐľ ľ ; 5Nę Lii ,Ŵ Li
+=
L ľ ! Li DI
ę ľ ę&ęę ľ ę 5=ę <
i Yi
Yľ
i ę"0ę i
"0ę ľ
ę
$ 5=ę ęęę $ęęę ę
ę
ę
ę <ę$ę$ę
ęÆę
ę ©ę ę +ę44ęę + ęĔę
ęfę
9ę
5=ę ę ę aę#ę,ęľę ę8ę Qv ,ęľ b ę8ęę ľ
ľ
; 5=ę ;7 6ę#ęuę3ę3ęęęę:ęę 0ę 'ľ ę`ęľ E7ę6ęę#ę ęękę#ę3#ęę#3e ęęę ľ 8 ę;7 BI
9 5=ę aę#"ęęę("! Ìę +ľ ę ľ l H[i +ľ H]ię H^ię H`ię Hcię9ę ľ 9 l! ę ę
5=ę ),)ę ę ę)ę yę 7ę 6ęďę "ęęIę 3#ę(ęę !ę AęAę!ę !ę9ę 6ęI ę 3#ę ę&ę !ęAę;ęęę 6ęęę
Ięęę i ę ię
DI DI DI
5=ę 'ľę; ę DI
,; 0MMot ęyę; ę ľ ;ęľ HŴęęHŴ 6ę
ę + ę'~ľ
5=ę T3ę##ęęHFę y##ęę£¤S?H!ęSFBęB?H7ę T#(ęSFBęBRD7 â(ęę#ę#!ęSRBęB?DBęBHFę# ?SBęBHRBęBFD7ę 6ę~SDRAę~DHFę# ę~H?Sę# #ę##"#ę# ę#3ęęüę#ę##ę7ę 6ęę#ę##ęęęk#"ęę g 5=ę 7 {#(!ę ę ęę ę ę ę ę 3ę ¦Ŵ }#ę ę ę ?Î! ? 8 !ę NŴ ę 33e& 3gę ę #ę ę&ęę ęę ę#ę# 7ę &!ę ę#(ę&ęę j\K N0KŴę? M!ę ę #"ę í0ĥŴ 8i ę 6!ę )ę 0ľ ę @ę)ęęAę ę #eęęę ęę+3ęę#ę #ę"#ęĕ#"²ę ęę&ę#(ę ę 3 7ę 6! )ę ę7 +
F
F
I
5=ę a"ęę#3ę#ęÐ,Ŵ qAę &ę"ę`ęę.`ę ľ B ;ę ,Ŵ &ę`ęęB ;ę 6ęęę B7
ę
ęRŴ # ę ę(
$ 7>K $ Ię#ľ K Ięğ#ľęę K <ę &7#&K ?|ę =ę |ęę?Dę Nę Dęęę =ę ę ,@>K ?|ę =ę ?ęę?DęNę ?ę ę 7 ę ę!7%K Ięā;ľ ę =ę $ę &7#&K Ię;ľ $ 7>K
$ &?K ]ęę
ľ ę i $ę ,&7K ]8 t
9ęę]<ę 84C&K ?8K )7%K?,&K98>0?0C&K ;88?K ]ęę;7ę &7#& 8ę ľ ę ľ
ęę !ę >8K?,&K 9;8%@#?K 8'K ?,&K ;&!4K ;88?>K0>K
$
v 7>K < ">&;D&K?,!?KE&K"&*07KE0?,K ę904&>K8'K >?87&>K !7%K&7%K@9KE0?,K ęi9ęi9ęi ę ęę904&>K8'K 87& >?87&> K ?K ?,&K &7%K 8'K &!#,K ?A=K ?,&;&K 0>K &F!#?4HK 87&K 58;&K 904&K 8'K >?87&>K ?,!7K ?,&K "&*07707* 8'K ?,&K ?BK ,@>K ?,&;&K #!7K "&K &G!#?4HK ęm
ę ľ ;ę 4&*!4K ?@=>K &7#&K 4!H&;K ę 0>K ?,&K );>?
94!H&;K E,8K #!778?K 5!3&K !K 58C&
;$ 6K $ 07#&K :ę ÌÍŴ )ę!7%K:8ę
)Węę2)ę :ęQAę E&K +&?K :ę ľ )ę$ę ę ,@>K :8ę ľ )Węę2:ę ľ )Qę ľ ęęę !7% :)ę ľ æã:ę ,ľ )QWę :8ę )8 ęęm$ę ,@>K :) : Eę · )Eęę:) 2:ę ľ )Q2:8ę ľ )8ę :)Qę ę$ 2 2 2QQę
9$ 7>K 9$ 07#&K
8 ę ę §ę ¿ !ę ?,@>K ę!ę!ęAę;ę ,&0;K >@5K0>K 9vę ę
v 7>K $
]0ę '8;K >85&K 98>0?0C&K 07?&*&;K ] 0504!;4HK ęę
Wę '8;K >85&K 98>0?0C&K 07?&*&;K $ę ,@>K ]ę Rľ ęę $ę ,&;&K !;&K ę 9!0;>K 07 ?8?!4K 7!5&4HK 2 !ę Q!ę2!ęQ!ę$ $ $ ęAę2 Aę Q$ 07#&K ęę
ę ľ .ęm
K 7>K
ę
0>K !7K 07?&*&;K .ę ľ
$
07#&K ?,&K 9;8%@#?K 0>K !?K 58>?K !K ?-;&&%0*0?K 7@5"&;K ?,&K 98>>0"4&K !7>E&;>K
!ę !ę ę 8;K $ę 07#& ę ę Ŵ ľ :Ŵę Ŵ ľ ę ę ę ľ ęę ę ľ ęęę ľ ęę:Ŵ ľ º Ŵ ę ę ľ ę ę ę ę ľ ęę ę ľ ęęŴ ľ
ę ę ľ
ę ę Ŵ ľ Ŵ ę ę ę ľ ęę ę ľ ę ę ęľ ę ę ęľ !ę ę ę ľ ęęŴ ľ Ŵ ?,&;&K!;&K ę 98>>0"040?0&>
Í ę
Uę >,8@4%K *&?K 0>K
I
7>K &?K @>K @>&K ĉ>ę ?8K %&78?&K ?,&K 4&7*?,K 8'K 87&K >1%&K 8'K >:@!<&K ę ,@>K ĊR>ę K
/nę &?K >ç>ę "&K ę !7%K >D>ę "&K 1ę ,&7K >S>ę 1>K ę/Kę>H>ę 1>K ęLę>è>ę 1>K ę ę/Kę>å>ę 1>K ę ę/Kę>F>ę 1>K ę ę1ęę/Kę !7%K >?>ę 1>K ę ę 1ęę/ę &7$&K E&K ,!C&K , ę ę 1ęę/ę , ęęK , ęę1ęę/ęę, ęę/ęę, ę ę/ę , ę ę 1ęę/ęę, ęę1ęę/ęX, ęę , ę/ęę, ę ę/_ę Xęę !7%K 1ęK nę &7$&K?,&K !<&!K8'K ?.&K<&$?!7*4&K 1>K ę)ľ ęXę ę
,&<&'8<&K
7>K 8?&K?,!?K
1ę ę ę !7%K ,ę 1ębi ę ęę \ęę1 \ęęę ę ÀXę á_ ęę \ę ęę1 \ę ęę \ęę1 \ęę,ę 1ę\ę ę
,&7K
ęK 1ęK Lę E&K*&?K ę ,@>K ?,&K >5!44&>?K C!4@&K 1>K gę
7>K g )Eęę
ęK ,)ęę ,)0 )ęę / / nę 8K1'K )ęę ę %1C1%&>K )Eęę
Lę)ęę ę 5@>? %1C1%&K / ę ,@>K?,&K4!<*&>?K )ę 1>K _
n 7>K 0 V[ ę /&7K 0 V[ ę K . Lę 0 V[ ę K . ę ,@> ę ę ę ę K ę ,&<&'8<&IKK Åę 8IK 0V ęK ę h3 0 ęX .ę . ę . ę
,&<&'8<&K ę(KK g &?K .ę K
$ 7>K Eęę1.EęK , 0ęę1.0,ęę. , ę1..ę ,@>K E&K *&?K ęK /,ęę. Ę_ę 2514!<4H ę ûýęK , Eęę1.E,ęę. , 0ęę1.0._ę 8K E&K *&?K ęXV ę ,ęę.ę /ę 84C17*K E& *&?K ęę.ętę !7%K ęK9ę ,&<&'8<& ę oęę1 āę K , ęę1 Ă,ęę. , Eęę1ăęK , ,9ęK ę
ę
2;,#@>B)Y #E/)8#E2Y >&2)EHY !& # # & % & & 8 "Ŵ $!& #& & $& 5 !ZĨ [Ĩ£Ĩ 7Ĩ
q.À .Ĩ
!# #& JJ1?CJQ +GQ ?+@OQ DK1GJ6B@GQ +GQ OBKQ -+@Q &BQ -+:-K:+JBFGQ +F1Q +::BM1/ Q )5BMQ +::Q J51Q GJ1CGQ 6@Q OBKFQ MBF96@4 Q +-5Q DK1GJ6B@Q -+FF61GQ 5Ŵ ?+F9G Q
$ 1 Ĩ Ĩ Ĩ G YOQ Ĩ :!ĨĨ U NQ >ĨOQ Q
GYR "2Ĩ,ĨO $
$ #Ĩ : Ĩ TY Ĩ Ĩ y Ĩ Ĩ Ĩ Ĩ :Ĩ Ĩ
Ĩ Ĩ: ÑĨ SY >Ĩ U 5 ĨĨĨ: Ĩ Ĩ*Ŵ +i FĨ ĨĨ ZĨ Ĩ Ĩy Ĩ ĨĨ :Ĩ Ĩ
Ĩ Ĩ: $ .$ 5 Ĩ Ĩ Ĩ :Ĩ ľ Ĩ Ĩ H !Ĩ Ĩ Ĩ :Ĩ : ZĨ Ĩ Ĩ ľ Ĩ Ĩ
Ĩ M!Ĩ Ĩ Ĩ Q Ĩ Ĩ Q Q Ĩ Ĩ ! Ĩ Ĩ Ĩ Ĩ Ĩ Q Ĩ hĨ ĨĨ Ĩ :Ĩ Ĩ Ĩ Ĩ :Ĩ Ĩ Ĩ: Ĩ! Ĩ Ĩ Ĩ ! r [ _Ĩ Q Ĩ Ĩ :Ĩ "Q Ĩ CĨ Ĩ Q Ĩ Ĩ Ĩ :Ĩ "Q Ĩ Ĩ Q #Ĩ Ĩ Ĩ Ĩ Ŵ Ĩ 'Q Q ' Q _:Ĩ Q ä Q Ĩ "Q 7©$ !Ŵ
Ĩ Ĩ Ĩ Ĩ Ĩ :Ĩ Ĩ U Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ
! ĨĨ ĨþĨ ľ :ĨĨ Ĩ ĨĨĨ yĨU Ĩy Ĩ $\Ĩ Ĩ Ĩ Ĩ Ĩ©k$Ĩ !ZĨ ľ Ĩ Ĩ!Ĩ Ĩ $Ĩ # uĨĨ: Ĩ ľ Ĩ7Ĩ Ĩ Ĩ.$Ĩ 5 ĨĨĨ Ĩ Ĩ: ĨĨ[Ĩ $ Ĩ
2;,#@>B)Y #E/)8#E2Y >&2)EHY !4<.$A?C*Y $F0*:$F4'$7Y 7I:A4$(Y !Y yssä $ & #& & $& $#"&
5 Ĩ Ĩ2-Ĩ Ĩ Ò
ª
2 #Ĩ ."3 kAĨÓÔĨ [Ĩ $$Ĩ
3 kAĨ,Ĩ[
"Ĩ, Ĩ ĨÛä Ĩ
Ĩ Ĩ 0Ĩ ,Ĩ$Ĩ 3 0 "ĨV
ÎQ
#Ĩ"Ĩ Ĩ-ĨĨ Ĩ"-Ĩ2=Ĩ
$Ĩ
Mv
Íä
Ĩ' ĨĨk-Ĩ-Ĩ' Ĩ-Ĩ'Ĩ-ĨkĨ-Ĩ'Ĩ Ĩ ĨĨ $Ĩ
ç ĨĨ SY,Ĩ1Y Ĩ RJY 5 Ĩ SY Ĩ 1Y ěĨĨ Ĩ Ĩ Ĩ Ĩ Ĩ 8ĨR : Ĩ Ĩ xĨ 'i EDĨ Ĩ Ĩ Ĩ Ĩ Ĩ xĨ 'i E$Ĩ ·5 Ĩ DY Ĩ QY,Ĩ Y Ĩ Ŵ8Ĩ9lĨ # hi ĤŴ,ĨJCŴ 0Ĩ ŭCŴ # Ĩ 9Ĩ0Ĩ FĨ $$Ĩ 9ĨĨĨ ĨĨ Ĩ Ĩ Ĩ Ĩ
Mv
lv
gv
hĨ ĨĨĨ Ĩ Ĩ Ĩ Ĩ Ĩn3 lĨ `ĨĨĨĨ Ĩ ģĨ nĨV Ĩ Ĩ Ĩ !Ĩ Ĩ ĨĨ ĨĨ Ĩ Ĩ ĨĨ Ĩ Ĩ ! !Ĩ ¯Ĩ Ĉ: $Ĩ#ĨĨ ĨnĨV Ĩ Ĩ $Ĩ ´
ayĨ Ĩ ĨĨ Ĩ ĨĨ Ĩ n 3 Ĩ \Ĩ 5 ĨĨĨ ĨĨUĨ ĖĨ ĨĨ Ĩ! !Ĩ \Ĩ `Ĩ Ĩ ĨĨuĨ ėĨ ĨĨ !Ĩ Ĩ ¦ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ę Ĩ Ĩ Ĩ xĨ \Ĩ #Ĩ Ĩ Ĩ Ĩ Ĝ Ĩ nĨ3 Ĩ Ĩ Ĩ Ĩ ĨĨ Ĩ Ĩ Ĩ Ĩ Ĩ *Ŵ Ĩ Ĩ Ĩ Ĩ Ĩ ü !Ĩ $ĨB ĨĨ ĨĨ ĨĨ Ĩ Ĩ ĨĨxĨ Ĩ!ĨĨ Ğ Ĩ Ĩ !Ĩ ĨĨĨxĨ $Ĩ #Ĩ Ĩ Ĩ Ĩ ĨĨn Ĩ Ĩ3 kĨ \Ĩ
Y
93Ŵ
ľ
BĨĨ &Ĩ8Ĩ<6Ĩ `Ĩ T WĨ KDĨ T Ĩ Ĩ 9DĨ TdĨĨ ïĨ Ĩ T" > 9ĨĨLzĨ Ĩ"Ĩ>Ĩ9Ĩ>Ĩ ľ ĨLz%Ĩ #ĨKĨĨ 9Ĩ>Ĩ ľ ĨĨT dĨĨ KĨĨ 7z%Ĩ
BĨĨĨĨĨĨ ĨĨ%Ĩ aĨĨĨĨĨU ĨĨ Ĩ U Ĩ Ĩ Ĩ Ĩ"Ĩ Ĩ Ĩ>Ĩ SĨ>Ĩ ľ ľ ľ >ĨĨ"Ĩ3 'Ĩ Ĩ"A6Ĩ #Ĩ Ĩ
!Ĩ H ĨĨĨ ĨĨĨHĨĨ ĨĨU ĨĨĨ Ĩ Ĩ!;6Ĩ #ĨHĨĨ Ĩ !Ĩ ĨĨĨĨU Ĩ Ĩ Ĩ Ĩ Ĩ KDĨ9DĨ%Y Ĩ;8 Ĩ ĨKĨ>Ĩ9ĨKĨ>Ĩ Yľ 9Ĩ>Ĩ .ľ BĨU Ĩ ĨĨ ĨĨ N AĨ Ĩ" %Ĩ #Ĩ ľ ĨĨĨ Ĩ !Ĩ ·
ľ
Ĩ;AĨI KA I 9A I
%Y
ĚĄ Ĩ ;Ĩ8ĨK > Ĩ9Ĩ K > Ĩ Xľ 9Ĩ>Ĩ ľ
#Ĩ;ĨĨ !1Ŵ KĨĨ SĨ 9ĨĨ %YĨ+Ĩ#Ĩ ľ Ĩ4Ĩ ĨĨĨ % Xv X Xv [v`v Xv \v Yv [v v gv gv Xv ]v Xv ^v Xv _v
5I
gv
%ľ &ľ
&ľ
v##
`v [v
?%ľ ?ľ #v
&ľ
%ľ TUVĨklmľ
aHĨ Ĩ Ĩ Ĩ ľ Ĩ oĨ Ĩ Ĩ 6Ĩ `Ĩ KĨ *i 9Ĩ *i .ľ 1Ĩ uH Ĩ;Ĩ Ĩ Ĩ u Ĩ Ĩ Ĩ Ĩ Ĩ |ľ 1Ĩ Ĩ Ĩ Ĩ !Ĩ Ĩ Ĩ KĨ>Ĩ9ĨĨ;Ĩ3 %Ĩ #Ĩ 9ĨJi ; I ' m%Ĩ fĨ K / 3 PDĨ Ĩ 9Ĩ ľ ĨPĨ Ji ; ' m%Ĩ # ľ
A AĨ *i;A I PA I ; I I P' Ĩ Å i I ' I SP I P; I ' ' l
5ĨĨ Ji PĨ Ji Èi3 Ĩ ' mĨĨĨĨĨHĨ ĨĨ SPA3 P A ĨPĨĨĨĨPĨĨÆ ľ *i 4;I 7 ĨWĨĨPĨĨ'Ĩ Ĩľ *i ; > 7;3 S' m4Ĩ Ĩ Ĩ PĨ Ĩ ; 3 ' m%Ĩ #Ĩ ľ )i 7Ĩ ;Ĩ Ĩ SZĨ ľ )i Ĩ Ĩ;Ĩ Ĩ 4 Ĩ ľ *i S ĨĨ;Ĩ Ĩ !1 ľ (i LĨ ; Ĩ 7%Ĩ 1;Ĩ (i EĨ ľ (i %Ĩ #Ĩ Ĩ !ĨĨĨĨĨĨ iĨ ÇiĨĨ SĨ ĨĨ K Ĩ9ĨĨ %YĨ%Ĩ 1; Ĩ 4 Z ĨĨ Ĩ cŴ 9DĨ 'Ĩ ĨDĨĨ' DĨDĨĨ'%Ĩ #Ĩ ľ Ĩ7ZĨDĨ S6Ĩ #ĨĨĨĨ%Ĩ 4Ĩ
aÐĨ ĨĨĨ Ĩ 4Ĩ Ĩ Ĩ !Ĩ Ĩ Ĩ HĨ Ĩ 4Ĩ 6Ĩ fĨ Ĩ ĨĨĨĨĨĨĨ Ĩ!ĨĨ$Ĩ ĨĨĨ ĨĨĨĨ Ĩ Ĩ Ĩ %Ĩ fĨ Ĩ ĨĨ Ĩ.-Ĩ Ĩ Ĩ Ĩ Ĩ 4Ĩ 6Ĩ fĨ 1$Ĩ Ĩ Ĩ Ĩ Ĩ ĨĨ Ĩ.Ĩ ĨĨEĨ ĨĨĨ Ĩ ĨĨ Ĩ6Ĩ aĨĨ Ĩ !Ĩ -ĨĨEĨ ĨĨĨ ĨĨĨ ĨĨ6Ĩ fĨĨĨĨĨ Ĩ HĨ Ĩ Ĩ ĨĨ Ĩ S%Ĩ fĨĨ ĊĨ ĨĨĨ Ĩ -Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ 7Ĩ Ĩ1Ĩ '$ĨfĨ ĨąĨ Ĩ Ĩ Ĩ Ĩ -Ĩ -Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ .Ĩ Ĩ1Ĩ.=Ĩ #Ĩ !ĨĨ ĨĨ -Ĩ Ĩ Ĩ -Ĩ -Ĩ ĨĨĨĨ Ĩ1Ĩ 4 =6Ĩ #ĨĨĨĨ 6Ĩ 'Ĩ
[v
7vv* *I v56v6v* : *Ii:vZv89v"v
=Ĩ
.=Ĩ
'Ĩ
FĨ
4 =Ĩ
2;,#@>B)Y #E/)8#E2Y >&2)EHY & # # & % & & "Ŵ & #& q.¾o Ĩ
# !DĨ .Ĩæ!Ĩ7Ĩ # #& @GM1FQ $$Q ³¶Ŵ DK1GJ6B@G Q @J1FQ OBKFQ +@GM1FGQ B@Q J51Q +@GM1FQ G511JQ CFBL6/1/ Q
BFQ J51Q ?K:J6C:1Q -5B6-1Q DK1GJ6B@GQ 1@J1FQ OBKFQ +@GM1FGQ 6@Q J51Q +@GM1FQ G511JQ ,OQG5+/6@4Q J51Q ,K,,:1GQ -B@J+6@6@4Q J51Q :1JJ1FGQ B¶Ĩ
:ä JDĨ ä BFQ M'Ĩ -BFF1GCB@/6@4Q JBQ J51Q -BFF1-JQ +@GM1FG
BFQ J51Q BJ51FQ G5BFJQ DK1GJ6B@GQ MF6J1Q OBKFQ +@GM1FGQ 6@Q +@GM1FQ G511JQ +@/Q G5+/1Q J51Q +CCFBP CF6+J1Q ,K,,:1GQ ,1:BMQ OBKFQ +@GM1FG Q &BQ GJ1CGQ +F1Q @11/10Q JBQ 8KGJ63OQ OBKFQ +@GM1FG Q +-5Q DK1GJ6B@Q -+FF61GQ Ŵ ?+F9 Q &BQ -+:-K:+JBFGQ +F1Q +::BM1/ Q
l `ĨCQ
Pv
pQQÎĨ DQ
3I
0Ĩ DQ 0ĨFQ
B'Ĩ
CQ
M'Ĩ
ĻŴ 0ĨFQ
.AQQĨ ĨFQ
6ä
CQ
1I
FQQp lĨ hY YĨ ĨYĨ ĨĨ Ĩr
0ĨFQ 0Ĩ D
°J'Ĩ
DQ
0ĨCQ 0ĨFQ
3ä
FQ
0Ĩ CQ 0Ĩ D
0ĨCQ
$ h YĨ ĨĨ ĨĨĆĨYĨr B'Ĩ .AQĨ
6ä pQ
J'Ĩ . QĨ>ĨAQĨ
4ä .ĨŴ o0 AQĨ
M'Ĩ .Ĩ ľ 1 Q
.l hYĨ Ĩ YĨ Ĩ
ĨĨYĨ
B'Ĩ Ĩ
6ä [
J'Ĩ 7Ĩ
3ä L
Ĩ [ `tĨ NQ tĨĨtĨYĨĨ NQ > P Ŵ
B'Ĩ [LĨ
6ä F
J'Ĩ FĨ
M'Ĩ Ĩ
Pv
Ĩ [$Ĩâć ĨtĨtĨĨ N Q >Ĩ ² Ŵ
2ä F4 7Ĩ
M'Ĩ aĨĨ YĨĨ
. Ĩ Ĩ JĨ + F+ J Ĩ Ĩ Ĩ )Ĩ Ŵ NQ Ŵ 7NQ Ŵ oĨ Ĩ Ŵ ŴcĨ 1 ĨĨ)Ĩ Ĩ Ĩ ĨĨ ĨĨ Ĩ)÷Ĩ+Ĩ 1I
B'Ĩ Ĩ
] .
Q
1I
J .ä .Ĩ 3ä Ĩ M±Ĩ F
7+ Ĩ Ĩ Ĩ v)DĨ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ ġĨ Ŵ Ĩ Ĩ Ĩ Ĩ 4Ĩ bDĨ dĨ .Ĩ bDĨ T <Ĩ T< &$Ĩ 5) Ĩ Ĩ Ĩ Ĩ ) Ĩ dĨ<Ĩ Ĩ NQ AĨ ĨĨĨ Ĩ Ý
1I
2I
OQ
B'Ĩ EĨ
] .ä LĨ
J.ä 4Ĩ
4ä F
M=Ĩ 7Ĩ
ÚŴ
E6 1Ĩ) ĨĨ ĨdĨ<Ĩ Ĩ&Ĩ ĨĨĨ ĨČĨĨēĨĨF¼ Ĩ !Ĩ Ē 6 ĨĨ Ĩ)ĨĨvĨ Ĩ Ĩ<Ĩ Ĩ)ĨĨ Ĩ Ĩ Ĩ!Ĩ!ĨĨ ĨĨ Ĩ)ĨĨ !r B'Ĩ E4Ĩ
] .ä E7Ĩ J .ä ELĨ
3ä LĨ M=Ĩ L
L+ #Ĩ Ĩ ľ Ĩ ĨĨ HĨ Ĩ ĢĨ Ĩ Ĩ) Ĩ ĨDĨ.DĨľZľ ľ !6 èĨ Ĩ Ĩ § Ĩ Ĩ Ĩ Ĩ Ĩ HĨ óĨ Ĩ )Ĩ )Ĩ Ĩ Ĩ )Ĩ Ĩ )§Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ HĨ Ĩ F4 L6Ĩ _Ĩ Ĩ )Ĩ Ĩ Ĩ Ĩ ĨĨ HĨĨ Wľ vôĨĨ )Ĩ Ĩ yľ
/ä oĨ
] .
Q
J .ä .Ĩ
4ä 4Ĩ M=Ĩ FĨ
EĨ
q% 5 Ĩ gŴ fŴŴ Ĩ Ĩ Ĩ )Ĩ Ĩ ŴŴ Ŵ .Ŵ Ŵ Ĩ Ĩ Ĩ Ŵ Ĩ 3 $Ĩ Ĩ Ĩ Ĩ "ĨĨ .Ŵ r / O %Ĩ hĨ ĨĨ ĨĨĨr Ŵ
Ŵ
B'Ĩ "8Ĩ
Ŵ
]=Ĩ " Ĩ J=Ĩ V 0 " 0
5ä V 0 " 0 V
M=Ĩ "Ĩ0Ĩ3 Ĩ Ä Ĩ Ŵ Ĩ Ŵ Ĩ ăĨĨĨĨ Ĩ Ĩ º3 V Ŵ
Ŵ
Ĩ % Ŵ Ŵ
c 1 ĨĨĨ Ĩ Yâä /Ŵ PY 6 Ŵ
B=Ĩ 4Ĩ
]=Ĩ FĨ
Ŵ
J=Ĩ 7Ĩ
5ä kĦ M=Ĩ L
% 1 ĨĨďĨ
?7cĨV qq4 cĨ 7Ĩ
?% 1 ĨĨĨĨĨ ľ ĨvĨĨ!
.% 1 ĨĨĨĨĨ ĨŴ Ò}iä 4 +Ĩ1 ĨĨĨ ĨĨ Ĩ ĨĨ "Ĩ2- ĨÿĨĨ) ?"ĨŴ 72Ĩ Ĩ?7%Ĩ
F+ ª«¬ CĨ ĨĨ½ ĨĨĨå CĨ Ĩ újĎę BĨĨCĨ Ĩ ĨĨ Ĩ Ĩ Ĩ Ĩ Ĩ CcĨ Ĩ Ĩ Ĩ CĨ Ĩ 6Ĩ #Ĩ Ĩ C ĨC cĨ ĨĨĨ Ĩ Ĩ Ĩ%Ĩ _Ĩ ĨĨFĨbĨ CĨ Ĩ?ĨbĨ Ĩ ĨĨ"Ĩb ?4 Ĩ v ĨĨ)Ĩ 6 'Ŵ
ÏŴ
LĨ
7% M Ĩ cĨ
Ĩ Ĩ Ĩ Ĩ Ŵ Ŵ Ŵ Ŵ pĨ? ¡ä Ì? ¡ä ÍĨ? ¡ä Ĩ? ¡ä
E% _(Ĩ Ĩ
Ĩ î -Ĩ C<Ĩ *Ĩ Ĩ *UĨ #Ĩ (*Ĩ -Ĩ ( Ĩ~Ĩ Ĩ ā(% #Ĩ (Ĩ*(*ĨCĨ ( Ĩ<~Ĩ (*Ĩ &-Ĩ ( ĨĨ(Ĩ*(*Ĩ<~Ĩ ( Ĩ C
(*ĨĨ ^$Ĩ 5 *ĨĨ<& /Ĩ|Ĩb-Ĩ&^Ĩ /ĨĨb-Ĩ ( Ĩ^~Ĩ Ĩ"Ĩb%Ĩ 1 ( Ĩ Ĩ Ĩ"%
=Ŵ
L% 1 ( ĨĨ*Ĩ ĨĨ ý (*Ĩ ĨĨ !(
% (Ĩ * Ĩ.
Ĩ (*ĨĨ Ĩ Ĩ ĨuĨ
*ĨĨ ?ZĨ .-Ĩ |-Ĩ E-Ĩ (
(*Ĩ ĨĨ
*ĨĨ %Ĩ B*Ĩ(ĨH-Ĩ *Ĩ * Ĩ.
Ĩ (*
( Ĩ.|?-Ĩ |EE-Ĩ-Ĩ %Ĩ 1 ( ĨĨ*ĤĨ úĨĨĨ
* (Ĩ * Ĩ.
Ĩ (* Ĩ (Ĩ *Ĩ !% ?% 1 ( Ĩ ĨĨ
Ĩ N 4Ĩ* (ĨEz%Ĩ * (ĨzĨ
Ĩ ?% Ĩ (ľ ĨĨ áäŴ ÖäŴ ØäŴ ×oä 1 ( ĨĨĨ Ĩ Ĩ v , pÊ% (ľ
??% 5 *ĨĨ( Ĩ ĨĨ Ĩ(*Ĩ*Ĩ * (Ĩ,Ĩ* ( ĨĨ ( Ĩ *Ĩ,Ĩ * Ĩ/Ĩ%Ĩ 1 ( Ĩ ĨöĨ Ĩ?Ĩ *Ĩ?ĨŴ 4Ĩ *Ĩ? %Ĩ Ĩ
% U Ħ ĦHĦ
%3, %3, %3,Ħ55Ħ %3,Ħ D Ħ@ Ħ ľ ĦĦ ĦĦľ vĦĦ Ħ Ħ Ħ Ħ Ħ ĦĦĦ Ħ 'ĦĦ Ħ 'ĦĦĦ %3,ĦIĦ Ħ ĦĦ ľ,Ĩľ
35 ĦAĦ Ħ 9Ĩ ĦĦ Ħ ; ĦĦ #Ħ Ŵ A#Ħ,Ĩ9ĨĨ ĦIĦ Ħ ĦĦ 9¿ AY
3ÝËĦĦ Ħ Ħ Ħ Ħ <
,5 ; Ħ#ĦĦŴ Ħ Ħ Ħ #$ 3& #z S&Ħ/Ĩ V5Ħ IĦ Ħ Ħ Ħ ĦĦ#Ħ,Ĩ Ĩ
S5 EĦ ĦĦ ' C /ĨT ¥yĦ Ü&ċ C -Ĩ & /ĨC Ĩ < Ĩ xä 'Ħ <&,Ĩ CĨ ĨĨ '5Ħ; Ħ CĨ ĨªrÜä IĦ Ħ ĦĦ# <Ŵ
ÙŴ
k E nā ª*ä Ĩ
Ĩ ,Ĩ ,Ĩ, Ĩ Ĩ
Ĩ,Ĩ
ĦĦ Ħ Ħ#Ħ@Ħ Ħ ĦĦ û=Ĩ,Ĩô %&ĦŴ ,&Ħ,Ĩ $6Ĩ,Ĩ
ä
7ŘŴ ZŴ Ŵ
!
Si
+mä
H EĦ Ħ@ Ħ ĦOĦĦĦ¢ĦĦ ¤ĨAŴ īćo©Ŵ ÓŴ ġPŴ Ŵ ķijŬĦn øĖ ľ @ŴAŴ çø Ħ Ħ ' \OĦĦ]sĦ ĦĦ ĦĖĦ øŴ äQŴ ĢPŴ ėģdmďĬŴ ïBŴ BôŴ ĚuĦ ¤ëĨĨ rO 5ĦE Ħ ĦĦßà\]rĦĦ,Ħ'ÈĦuĦ ŒęĐŴ łđŴ ĴĕŴ Ă@µùŴ ¾ŊŴ Ģľ ďėxľ ÙuĦ Ħ ĦĦ ªhä :I
ĶŴ
Ħ
0I
9 Ħ ā Ħ *ā ĦĦ (5Ħ ; Ħ ā āAā Ħ Ħ (Ħ Ħ Ħ "' Pāā*Bāā+ā IĦ Ħ ĦĦ
kä
+ EĦwĦ$2 Ħ$Ŵ
£M¥ä
1I
jä
0ā @Ħ Ħ ĦĦ wCĦ$2 CĦÓ+
ā
>Ħ'"Ħ( ( Ħ(ĦĦ Ħ (Ħªā%ā Ħ Ħ < ā Ýā
úāāêāā
1I
IĦ Ħ( ' (Ħ Ħ Ħ (
ľ
ľ
ā ľ EĨ ľ 'ā'ā'ā ľ ā ľ ā
Ħ Ħ"Ħ ā Ħ ĦĦ yĦ ( Ħ ?Ŵ Ħ Ħ ĦĦ (Ħ ' (yĦĦ ā ā Ħ Ħ ĦĦ(ĝĦ ' (Ħ; ĦĦ nā ā
9 Ħ <ā ľ ?Ŵ
)*v
Mv
ā
āā%ā
1I
āāā
ā
<āā āā ā %āā%āā%āā% %
(ĦĦ (Ħ ' (Ħā%ā IĦ Ħ ĦĦ nā ; Ħ 0ā. 0ā0ā'ā'ā'ā ĦĦ < ĦĦ ' (Ħ ĦĦÆ
1I
Çā Ħ
ā ā .āāāā'ā'ā'āā59. 5ā »6ā 1I
ľ
(ĦĦľ ,-i +ā IĦ Ħ ĦĦ ā ā ā Zè' ā āā āāāāBāā'ā'ā'āāā¹Ŵ' Ŵā#Ŵ · 6āā¸Zā ā āā + 9 Ħ$ā ĦĦ (Ħ ĦĦĦ(ĦĦ Ħ < ā $āāā$ā ā ā Mv
( Ħ Ħ (ĦIĦ Ħ ĦĦ $ +ā
ā
2;,#@>B)Y #E/)8#E2Y >&2)EHY 4<.$A?C*Y $F0*:$F4'$7Y 7I:A4$(Y yssä & #& $#"&
+ .Ħ
Ôā
ø!
ā Eā !
ā
!
āï ā îŴ ā ĹŴ ā
Qā ā ù!
ā ā !
ā ā !
¡Ħ !
+ā
Y
Mv
D ĦĦ¡Ħ 0Ĩ *Ŵ 0Ĩ ĺ + .Ħ ā ! āEā ! +ā ā ! āā! +ā ! āā āEā ! āā !Èā ā! āā ! ā 0Ĩ +ā ! +ā 0Ĩ ā ľ ! āā" ! +ā āā ā āā ! āEā ! ā ā ľ ! āā! + Mv
Ov
Mv
D 1ĦĦ Ħ ' ĦĦāā āā ! ā + .Ħ M
I ĦĞĦ ĦĦ ā ā ;)Ħ ā ĦĦ"Ħ 0ā BĦĦĦĦ ĦĦ +ā ;' "1Ħ ĦĦĦċĦ
ā Ħ ā DĦĦĦĦ
Qā Ħ 'ĦĦĦĦ
Qā ;Ħā ľ ā āā ĦĦĦĦĦĦĦ"Ħ Ħ ĦĦāĦ+ā ~ ĦĦĦĦĦ
Qā Ħā p1Ħ ā ā D ĦĦĦĦĦ āā āā
āā
Qā Ħ Mv
.Ħ U
ā
Ŵh ā Ŵ
ā ā āÊËľ ŴAā ā ā ā āā Ŵ
Mv
ā ā āā YŴ āāā K āLā Eā ā ā ā ā 56ľ ā°Ŵāā ¹āEā ā
56ľ
BI
Ŵ
ÀÁľ
ā
Ŵ
āā±Ŵ ā ā ā Ŵ
ā
.Ħ
Ħ@ĦB ĦB óBĦBĦĦB ĦBĦ 1Ħ Ħ ĦB Ħ < BĦ' Bģ "Y
1Ħ Ħ Ħ ā ā Ŵ ā Ŵ ā ā l Ŵ âãľ Òā -Ħ ā Ŵ āŴ ā ā ā `aā ā Ŵ ā-Ħāā ā Ĩ àáľ Ŵ ā ā dĦ Ħ āŴ ā8 ā `aāā ā g ā ā `a Av
³ä
ā `aāŴ āvā
tä
ĦB Ħ"ĦBĦĦB BĦĦ ~Ŵ
.Ħ
K ĦBBĦ ;#ā ;&#ā 1Ħ & āāā rā # ā#&ā > ĦĦ R8ßāā ā ľ ĦĦ;#8&ā ; Ħ8āā©ā&8āā& 8āā ā ā ā Ħ ĦĦ ;8āāā ľ ĦĦ;8&+ā āā ľ ĦĦ;#8&ā 1Ħ
ĦĦ8ā#ā ā ĦĦ Rā8ā Ŵ ĦĦ;8ā#ā āā ľ ĦĦ;#8&āŴ ĦĦ #8&ā ā kĦ ľ ĦĦR#8&Âā
ā
E\
.Ħ U DĦ ' ĦĦ"Ħ ?Tā ā Tā āTā ā ā ā ā āā ā," ā
"Ħ Ħ Ħ ĦĂĦ .Ħ U Ħ .Ħ Ĩ Ħ #ā ĦĦ Ħ Ħ@ Ħ ĦĦ 5Ħ EĦĦ 1Ħ Ħ ' Ħ
' ĦĦā ľ ā ľ Tā ā ā U Ħ.Ħ #ā Ħ Ħ@ Ħ ĦEĦĦ 1Ħ ' ĦĦ
' ĦĦ Tā ā ā U Ħ .Ħ #ā ĦĦ Ħ Ħ @ Ħ ĦĦ Ĩ Ħ ĦĦ Ħ EĦĦ 1 ' ĦĦ"ĦĦā ľ Tā ā" ā D dĦĦ ' ĦĦ"Ħ āā ā ā ā"āā ," ā " .Ħ ]
9 Ħ ľ Ħ Ħ ' ĦĦ Ħ 'ì Ħ5ĦD Ħ Ħ Ħā ä ľ ä 0ā ä
ā āā āā ā ľ ľ ľ āā ľā? " ā āā ľ ā? "ā ä ā ā ā ā āfā 0Ĩ 0Ĩ ? "ā ä ā ā ā ā 8Ĩ ā 0Ĩ ? "ā Ħ ? ā ä º ā ā ā ľ Íā ā ľ ā ľ ā ??āĦ ā?? āEĦ ā ?0ā ā ā ā ā ľ ā Ħ āā ā> Ħ ľ ā ? "āāwā ā
K ĦĦ
1I
.Ħ
K Ħ ā ā
*ā ā ā2f āā ā ā*ā ā2ā ā *ā *d
2ā f ā
; Ħ *2āL ā 0ā ĦĦĦ ā 0ā *āÙ ā Ħ2ā ā D 1Ħ ā ā*ā ā2ā8Ĩ0ā ĦĦĦĦ *ā *d 2ā ā0Ĩ āD Ħ ā ā
ā *ā
ā 2ā
*ā ā*dā ā2 *2ā ā
*ā *d 2ā ā 8Ĩā$ ā
.Ħ ]
/Ħ 9 Ħāā gä Ħ/Ħā ā> Ħ Ħ
ā
ā ā
ā
ā ā ~ āā/Ħ ā
/Ħ
M
ā ā
ā E ā āpā ā
Ħ àF
ā ā ā āā ÆÇľ āāā āāā ā ā ā ÄÅľ ā ā ā ā ā ÈÉľ
~ ā
Nv
Á Û ¤ ä üHMā ā
ā
ā
; Ħ ā Ħ/ Ħ 1ĦĦĦĦ ā8Ĩ ā Ħ`Ħā ā
āāÕ gä ā
=âāäā/Ħ=āā āā8ā=āā āāāÕäāHMā ā g
ā
ā ā HM ā ā HMā ā ā HMā HMā
āH ā Mā H{ ā ā ā ā ā þ
NĦ %
.Ħ
ā ā
āāH { ßä HM ā āāā
pā ā ā ā
- Äāā ā ľ - ā ā ā ľ ā ā ā ā
.
ā ľ
Qā ĤĥĦħľ ā ľ
; Ħ ā ā ā ľ ā Ħ ā ā Ñā8Ĩā Ħ Ħ Ħ Ħ ' Ħ ` Ħ`' Ħ ā "Ħ .
ā 0Ĩ
Qā Ħ ā
ā
lĦ -āāc)}ääā ācāāā ľ āāy
1I
,āā c ā
K ĦĦx ā ľ äľ ā ,āā ā ľ ā /ľ ,āā cā ľ ,āā ā ľ ā ā /ľ ,āā yāľ x ā
ÒÓľ
1Ħ Ħ' Ħ ĦĐ đĦĦ -āāÔ CĦĦ
Ħ āā%ā ä ā °±²ľ ñçā%
1I
ā
~ Ħ Ħ Ħ - ā- ā-, ' ā
ā - -, ā-ā ā
Ħ Ħ Ħ ā Ħ NĦ
Ħ Ħ Ħ 'Ħā Ħ)ā' Ħ)ā ĦN ā Ħ âNāā ä7 ā ĦNā ĦĦmj Ħ;' "Ħ Nā ĦĦ) ā Ħ N K āpĦ ãKāä)Þ ā ~ Ħ Kā )ā ā ā ā ªHhä E E G ä 1I
ā
Ħ
óā
ā
ā
ªä
1I
ā ā
ā ľ
ā
1I
ā
Å ā
NĦ ā
ā ā jā ā ā ā ā Ħ CĦiā ÍĦj ¼ā ÎĦiā ÐĦ ¸Ħāāà ¹ĦāāČÄÉĚñ÷Ħ āāºĦ āājµěĦ äjġÅÏĦā ľ ā ľ ā ľ ā ľ ā äj»Ħ, äā v Ħ- , ā ā
1I
1I
ā
, NĦ
9 Ħā %ā 'Ħ; ĦR#)/
RF/ ā Ħ Ħ Fā F % ÐÑľ Ni Fā 7Ħ ā F/ā
3Ħ Wā Þßľ
3Ħ
Kv
?I
Mv
#/ā ; Ħ R#)&ā RF&ā Ħ Ħ F %Fā 78ľ Mi 7Ħ Fā #&ā F&ā ā āWā 9:ľ ?I
Mv
%-ā ¬ä fä %ā ā
> Ħ Ħ Ħ %ā
3Ħ
Kv
%-āWā Wā %ā 9:ľ C
3Ħ ā
Kv
āāWā X ā !"ľ Wā 7Ħ
ā
" .Ħ
9
- āāB- āāI ā7Ħ. . ā'āā'ā'āā. āā
; Ħā ā ĦĦ . āò' ' 'ā .āā6ā 7Ħ- āāB- āā ā ā ā Mv
1I
.Ħ
p ' ĦĦ ¶Ħ Ħ ' Ħ7Ħ ā ,5Ħ K ĦĦ ĦĦ Ħ@ Ħā 1Ħā ā,ā Ħ ,Ħ' ĦĦ Ħ@ Ħ ĦĦ ĦĦĦ Ħ Ħ ' Ħ 1Ħ ' 7Ħ ,Ħ ľ -ā8ä Ħāāā āā,ā ľ ā ,Ħľ -āā Ħāāā āā,ā ľ ā ā ,Ħľ -āā Ħāāā āā,ā 1I
Mv
NĦ
ā Ħ7ā
ā
Áā ,āā 7Ħā āĦ,7āĦ7 āwĦ Ħ7ā ā äĦ7ā$ Ħ"7 Ħ7ā āD Ħ-āJŴ Ħ"7 ā Ħ7ā
ĦĦ"7ā Ħ7ā
Ħ7ā Ħ7ā kv
AI
1I
Ħ
JĦ
> Ħ Ħ Ħ Ħ Ħ ' lĦ Ħ 4 (ä Ħ =Ħ-Ħ4(ä Ħ 4Ħľ Ħ =$ 5Ħ =Ħä Ħ ã Ħ
Ħ
'I
lĦ S3 9 ĦÐä ľ ánä D ĦÐä iĦ D 1
(#= =Ħ-Ħ = =Ħ-Ħ Ħ-ĦÐä-Ħ(#=Ħ-Ħ $ Ħ Ħ-Ħ = $ Ħ(#= (#=Ħ-Ħ $$ Ħ = 0
=(#
Kv
ľ
(ľ
-Ħ
Mv
ýŴ
ľ
ċ
Kv
ľ
M lĦ H
D 1Ħ
Ð*ä
Kv
ä ľ S3
Ħ=Ħ-ĦĦ4Ħ Ĩ ÔÕ Ħ=Ħ Ħ$ Ħ4Ħ Ħ=Ħ-ĦĦ4Ħ ā Ö× Ħ=Ħ 2Ħ4Ħ Mv
Mv
Kv
Kv
Ħ= - ĦĦ=Ħ ¦Ħ ĦØÙ z Ħ4 ( Ħ-Ħ Ħz Ħ4&Ħľ Ö×Ħ4Ħ-ĦĦ$ ĦĦ4Ħ-ĦĦĦ4Ħľ Ħ Ħ4Ħľ Cv
ÚÛ
Ħ D dĦĦ=Ħľ Ĩ$ xä
Mv
Ħ
xlä D 1
ĦĦ =Ħ-Ħ3ĦĦ 4Ħ 2 ĦĦ=& -Ħ3 $ ĦĦ4 (ä Kv
HI
Mv
=7= -Ħ3Ħ Ħ Ħ=7=
Ħ Ħ2 Ħ HĦ
HĦ
H
G
% ÒĦ % pĦĦ ĦQĦ%Ħ-Ħ6ĦQĦ,Ħ 6Ħ ĦĦ ĦĦĦHĦ0!ĦĦ Ħ ľ 6Ħ H ĦĦ0ĦĦ Ħ Ħ0!ĦĦĦ Ħ< ĦĦ He %6,Ħ 6,, Mv
Mv
Mv
K ĦĦ6,,Ħ % Ħľ 6ĦQĦJĦ ĦĦĦĦ Ħ Ħ% ĦZĦĦ Ħ0! x %6,Ħ Ħ ĦĦ Ħ ĦĦZĦĦx Ħ ĦĦ ĦĦ Ħ0ĦĦx,§ ĦĦZĦĦ YĦ Ħ ľ QĦľŴ % Ħľ 6Ħ-Ħ Ħ % Mv
Kv
6Y .Ħ , I ĦĦĦ
6·ÊĦ0Ħ
Ħ#Ħb Ħ Ħ ĦĦ
2=#Ħ
ĦĦĦ
%bĦ
JĦ
#bĦ 6YĦ ? !ĦĦ0Ħ< ĦĦòĦõ ĦĦ Ħ!Ħ Ŵ
Kv
Kv
#Ħ-Ħ%Ħ
=6=
Fľ ĵľ
J ³ľ #Ì$ J#ĦŴ Ħ Ħ HI
# $ 6& #2%&ĦŴ Ħ ÉÊ Ŵ Ħ$ Ħ©9äB©eä 12ľ
!
6Ħ ?Ħ#ĦĦĦ ĀĦ Ħ #Ħ aä 0Ħ#Ħ 2fnYĦ PĦ#Ħ Ħ¨Ħ bĦ ±
Ħ ľ Ħ 6fnĦ $ fnĦ Ħ2 Ħ -Ħ^MĦ ?Ħ $hĦ PĦ #Ħ 2 Ħ bĦ Ħ Ħ 6Ħ8ĨĦ Ħ Ħ Ħ 6fnĦ e Ħ Ħ Ħ 0Ħ ĦĦ Ħ < ĦĦ Ħ-Ħ^8Ħ ĦĦ ĦĦĦ< Ħ Ħ! Ħ Ħ Ŵ$ e ^&Ħ2 ĦŴ ^&ĦŴ 6 *ā
$ ^& ĦQĦË'äŴ Ħ Ħ * Ŵ , Kv
Mv
Mv
Mv
Kv
Mv
v
Mv
, NĦ #$ 6& # S&ĦŴ VĦ12ľ #2 ,ĦQĦ& #$ ,$ &Ħ VĦ 34ľ #e ,&Ħ Ħ VĦ 34ľ #$ ,&ĦŴ Ħ-Ħ VĦ Mv
GI
1I
P Ħ ľŴ ā#2 ,ÂĦĦĦĦ ĦŴ ĦQĦ VĦ¾¿ľ VĦ
ľ
Kv
TĦ
Ħľ$ &Ħ Ħ ľ QĦ&ĦhĦ
1I
K ĦĦĦĦ< "ĦĦ"ĦĦľ T ĦĦ Ħ ľ Ħ0Ħľā Ħ ĦĦ! Ħ0ĦĦľ$ Ħ0ĦľāĦ Ħ ĦĦ ĦP Ħ ľ Ħā WĦ Ħ ľāĦā WĦŴ Ħ
X Wg_ĦācĦ Mv
?Ħ ľ ;Ħ W³_ĦāĦĦĦĦ!
0ĦĦĦĦĦ ĦĦ ĦU "Ħ Wg_Ħ Ŵ ĦUā Ħ Ħ Wg_ĦāĦā ĦØā 5ĦEĦ Uā 8ĦĦ Wg_ĦāĦUā Ħ0ĦĦĦ00ĦĦĦĦĦ ĦĦĦ ā Ħ Ħ ľ āĦ,Ĩ Ħā %Ħ Ħ
êĦ ā %$ c %ĦŴ u&ä Ĩ HĦÕĨ Ŵ ā %i
EĦĦĦ #2 &Ħ Ħā CĦāYā ā %ĦÖĨ#Ħ ĦÛä% ×Ĩ #Ħā HĦ Ħ MĦ Mv
Mv
ĦĦ !ĦĦ ĦĦ#ĦXŴ Ŵ Ħ HĦ ā%Ħ ĦĦ#Ħā ÁHĦ Mv
S¿ .Ħ % 0 Ħ)āĦ/ā ĦĦ)/ā E #& āĦ;#&ā0Ħ;)/ā Ħ !²!0 !iĦ oĦĦ& Eā )ā 0Ħ#& Eā )/ ā ĦĦ ;#&ā
Ħ /ā ĦĦ
Mv
>I
;)/ ā
Ħ
#ā ā ä 0Ħ V/ā)ā ā V) ā ?Ħ V)āāV)ā Y[\ å)āāV /)ā Y[\ ā
>Ħ æā/ā āY[\ >ā oĦĦ ;āàā /ā /ĨN Y 0Ħ /ā
Mv
)āā)/āā)ā#& ā
>Ħ #Ħ\ā
/ā /ā
Mv
Mv
Ħ
Ħ
» ØĨ #Ħā %M
Ħ ÓĦ 9Ħ &ā ā ā 0Ħ #&ā /Ĩ*ā ?Ħ Vā āY[\ā ĦĦ Ħ"! Ħ ĦĦ
%Ħ
?Ħ#& ā)ā 7ĦĦĦĦ)ā ā$ /ĦĦm#&ā m)/ ā ĦĦ =I
AĦ /Ħ R W ÔÕAĦĦ7Ħ/ 2/& AĦ $ /Ħ ! ĦĦĦ Ħ< "ĦĦĦ Ħ / 2/&+Ħ/Ħ Ħ 7Ħ? ÎÏľ /ĦRĦ 3Ħ 3Ħ >8Ħ jv
#&ā MY Ħ 7Ħ ä |ä !"ľ #Ħ7Ħ% ā#ā Ki
Ħ#Ħ 7Ħ
Ħ
J .Ħ , "ĦĦ"ĦAC2/CĦR A$ /& AĦ+AĦ /Ħ+Ħ/&8ĦĦ Ħ Ħ A +AĦ /Ħ+/Ħ Ħ
A2/Ħ Ħ AĦä/Ħ ACĦ$ /CĦ
9ĦAĦRĦ #Ħ+Ħ ĦĦ/Ħ7Ħ #Ħ$ hĦ Ħ ā Ui !ä
Ħ Ħý&Ħ+Ħ Ħ¿%ä Ħ¼$ä+Ħ Ħ½#ä ¾ A »ä #Ħ+Ħ &$ #$ &Ħ
7Ħ
Ħ 8Ħ &Ħ+Ħ< -ā+Ħ 55Ħ+Ħ TTJ&Ħ+Ħ TTT& RĦ Ħ²ä$ Ħ+Ħ´ä$ ²ä+MĦ ¼Ħ+Ħ·¸ä@ µ¶+ ä ĦÀä$ ¹º"
Ħ Ħ ĦĦ R® ľ RĦ¯ 7Ħ,Ħ
Ħ
8=
Dv
Ñ .Ħ ? čĦĦĦ ĦĦ ^KNGā Ħ Sā Ħ?ĦKGāā Gk0ā ĦĦ ĦĦ ^NGkā 7 ĦĦ ^KNGāRĦSāĦ
&Ħ
? ĦĦ ĦĦ KGlā ĦŴ ĦĦĦ Ħ ! ĦĦĦ ĦĦ mkGlā 7Ħ ĦĦ ^KGlā 7ĦŴ MĦ %Ħ
&Ħ
ÚĦĦĦ]OĦāðOĦĦĦ ā
ā
ĦĦ [OrsĦāxä ā ĦĦ []OsĦāËSā%ā
%&Ħ
9ZĦ ĦĦ ĦĦ [\]OĦāā ā ĦĦ [\ĦrOĦ ĦĦ
=
S% Sāā %ā
Ħ
3cĦ
Sā1ā % ā
Ħ ŏèòŴœŔŴ PQĴľūŴ
0FIiñ
4
=
VŗŴŴ
Ŵ ŎŴ
%
=îÿ:
ā ā)MĦ
ĦĦĦ
Sā%ā āā >ā
ā
I Ħā3cĦĦ 0ā ĦĦ %āā ā;ĦĦ ĦĦ[\ĦOsĦĦ ā )½Ħ .Ħ %
9Ħ$ā Ħ3ā ĦĦ ĦĦ ĦD ā āā āĦ$āā3āāā Ħ$3āI ā ā KĦ ĦĦ $ā Ħ3ā ĦĦ ĔĦĦ Pā*ā Bāāāā ā Ħ $Pā*ā $Bāāā1āā 3Pāā*3Bāāāā āI(ĦĦĦ< ĒĦĦ $3āāk¨`Ŵ ĦĦ $āā*ā3B ā Ħ 3āā*ā$B+ā ā $B3Bā āā ā$ :āā3Ā¢āā$āā3 $3ā ā$āā3:L$3$āā3ā :õā ā%Ħ
Ħ .Ħ ?)ĦĦ± ā ā ĦĦ) ĦĦ Pāā*āāā Ħ"Ħ Ħ ĦĦ Ħ Ħ2ā ĦX ā"Ħ) !ĦĦ))ĦĦĦ Õā XÚX 1ðwā ā ā .Q 2ā Ûā 2āā :āÖā āā ā ā ā ×ā āLā ā ā% ā 2āXāÜā ā ā2ā ā Xā ā ā
%ā
NĦ
CĦF$ CĦFĦ XùF$ ĦF&ĦXĦF+FĦ Ħ ĦċŴ+ĦĦ&Ħ £M ä jä ā+ĦwĦFĦĦF&ĦM Kv
/I
oĦĦĦ XF2{&Ħ
Mv
ā
ā
£ä= Êä Ħ
!"ľ
FĦ
ÜÝľ
ĦFĦĦFĦ
Lv
Kv
>ĦĦĦ CĦF$ CĦFĦ
/I
Kv
Éäā
Lv
qä
ā
ā
Ħ
+ +Ħ
$
¢>Ê ā
ā+Ħā éÌā ā
ā
ā
ā Ëä
áā Êä ā ā
((8Ħ
wä Æ3 CĦ{° CĦ{&Ħ ā Êä
Mv
Mv
ā ā ľ xä
Kv
z Ħ " ûā+Ħ%Ãāā
ā ā /I
ĤĥĦęĠĦ+ĦĜ+ Ħ Ħ¦§äāÌ2 ä ¦§ä [ ä Ħ¨U ä ā ÙÚÑ= ¨ä ā XĦ èĦ+ĦÓ+ Ŝŝ' þŴ ä ā @ā ľ ā ľ
`ĦĦ`Ħ(Ħ Kv
/I
Mv
āāāā āāāā ā ā ā ā@ā"ā"ā@ ā z NĦ @
9Ħ XĨ ā ľ .Ĩ ľ ľ ľ ľ ľ ā PĦĦĦ @ĦĦ((Ħ Ħ ëŴ Ħ X "Ħ ā K(ĦĦ ā "ā ľ ā 9Ħ @Ĩ
ā PĦ(Ħ!!ĦĦ@Ħ ((ĦĦ ìãŴ ĦĦ"Ħ " āK(ĦĦ ā+Ħā āā : :ā Îā :ā+Ħā ľ ā XĦ" ā Lv
Mv
Mv
Jv
Lv
ā
Ħ Ħ Ħ ā Ħ L:"Ħ ā : Ħ Ħ: Ħ L@Ħ ā ľ ā Ħ "
P Ħ @Ĩ
Mv
ā ľ ā ľ ľ ľ Įľ ľ ā ľ @ā ľ ā ľ
ĦL Ħā L Ħ:ĦĦ£ :Ħā ľ %Ħ ľ
ľ
ľ
ľ ľ ľ ľ ľ
BI
ľ ľ ľ
BI
ľ ľ ľ
>ā
+ā 8Ħ
ā Ŵā ľ ā ľ ā ľ ľ ľ ľ ľ ā . ľ %Ħ ľ Pā Ħ" ´ %Ħ Ħ"
?L "8Ħ ĦL Ħ ā L Ħv:ĦĦ£ :Ħ @ā ľ ľ ľ įľ ľ >ā 8Ħ @ā ľ ā ľ ľ ľ ľ ľ ā :Pā Ħ" ² ā Ħ" > 6I
>:8Ħ ā Ħ"
@Ĩ ľ ā ľ ā
E:Ħ Ħ 8Ħ @Ĩ
Mv
"ā
ĦĦĦ: Ħ āo8ĦĦLĦ X® /Ĩ ā ľ @ Ŵ ā ľ "āāā Mv
āā@ ā
8ĦĦ L: ĦĦ@ ā :.Ħ
9:ĦŴ /Ĩ0ā :Ħ <āāā
$=āÊā
$=
í£āā ā ]āā:āā > ā ā ā
:Ħ L
L:ú:8Ħ:ĦĦ ā ā
:ĦĦLĦ
<ā ā
ā ā
<ā @ľ <³ ā u ā āāÝä ā ā ā ā ā < < u ]ā ā ā ā ´ ā 9I
8I
7I
ā ā /ĨāāÞ< ā ā ā ā
!ĦĦ< Ħ! ĦĦ!Ħ <ā <ā ā ā
'
Mv
1I
ā āāā'āā'ā āā āāgā ā āLā ] ā ā
ā ā @I
Ħ
-=
ā ] ā ā āā
Ħ] ā ā ā ā
<ā ā ā 1I
1I
ā Cā
ā
>)8Ħ ā
Mv
ā ā
Mv
4 .Ħ
K aĦĦĦ ā 'i 5ā h 59Zā
1=
1I
6āā.āā'ā'ā'āā5sZā 6āāoāā'ā'ā'ā ā 59ý
Kv
ĦĦaĦ 5 h 59oā 59oā ÂÃľ 5ā Mv
1
=ÉāLāāā ā 59Z'ā
EĦĦ"ĦĦ))ZĦĦĦaĦ ĦĦaĦĦ ā 8Ħ
1I
ā|āā|āā ĦaĦ 5ā
ā 59Zā ā ā >ā ā ā 59 ā ā
K
1I
1I
1I
Lā4ā ā 44āā 6 āôā ā Lā ā 5 64
5
ā
>ĦĦĦ ā ā #Ŵ Ŵ 9 .ā Ŵ Cā ľ ľ İľ Ŵ ā .
Oā ā ā :ā ā ā äāOāCāā.āCāāāľ ľ ľ ľCā.
ā.
ā Bv
=.
āOā= = = = = ā = ' = = = 3444=.
ā t ā 9 OāD āOā Ŵ Oā Ŵ ľ ľ ľ {Ĩ :āOāD ā ā ā ā ā 5 > > ā : Ŵ {Ĩ āŴ 1ā ā ā āD .
ā ā ā ā
> NĦ JS
9Ħ āā ĦĦĦĦ! Ħ vĦĦĦ!Ħ< 5Ħ
C ā 1$rā $ā I Ĩ
ā1ā ā
Ħ!Ħ $ā ĦĦ ā1āāC ā āā ā 78ľ D ā1āā8Ĩ ā ÌÍľ _ā 8Ĩ §ā Ħ ā8Ĩ ā
? "ªĦĦĦ D ā8Ĩ ā «ĦĦĦ ā āāCāā D āā _āC $ ® _$āCāā 1ā ā =ľ 1āD ā{Ĩāā ā1āā =ľ ¿ā
=
=
P ĦĦĦ! "Ħ Ħ"Ħ ĦĦ ā 34i ā ?Ħ¢Ħ ;Ŵ (ĨEŴ 66;Ŵ (ĿĦ 8ĦĦ ĦĦ
D ā
1āā
ā 1āā
āD ā
1ā j ā_ā ā
Ħ @ ĦĦ Ħ< Ħ "Ħ ā ā 8Ħ ā 1ā, ā Ħ uĦ ňčőŴ ıĔŴ ĎļŖ(QĠIJĩʼn 0ĦĦĦ! Ħ 5Ħ >dĦ Ħ ĦĦ ā1ā ā ąĪEŴ ùEľ Ŵ åėē8Ħ .I
$ā }ā¥āā }ā , ā
ā
2;-#@>B)Y #E/)8#E2Y >&2)EHY & # # & % & & ¸Ŵ & #& & $&
? "Ħ ā Û Ħ ā
s ā
# #& II1?CIQ +GQ ?+@OQ DK1GI6BAGQ +GQ OBKQ -+@ Q &BQ -+:-K:+IBEGQ +E1Q +::BM1 . Q )5BMQ +::Q I51Q GI1CGQ 6@Q OBKEQ MBE96@4 Q +-5Q DK1GI6B@Q -+EE61GQ 5Ŵ ?+E9G Q
4 9ĦbŴ .Q Ħ*!Ħ )GĦGĦaŴ Ŵ.Q -Ŵ .Q ĦĦ*Ħ Ħ!ĦĦ4 qĦĦ .Q *ĦĦ ĦĦ4 4 9Ħ=4m|ĦĦĦ)"))Ħ< ĦĦĦ!Ħ)ĦĦ Q Ħ4ĦĦĦ& ĦĦĦ*Ħ !Ħ &ā ĦĦ*Ħm|Ħ*)Ħ =|ĦĦ}Ħ4mĦĦ @+
qĦĦ}=Ħ-Ŵ @Ĩ4Ħ }âÜ 1I
%5 Ħ)*)ĦĦ!ĦĦ)ĦGĦ*"ĦĦĦĦ G ā9ĦGĦ)GĦ =4Ħ GĦ!Ħ )*)ĦĦ!ĦĦĦą í Ħ )*)ĦĦĦ*Ħ %Ĩ qĦGĦGĦ* GāKā *)ĦÞ=æĦ4M 4 é ĦĦĦ!Ħ ĦĦ )! Ħ< *Ħ*! ĦĦĦĦĦ
" ĦĦĦ@Ħľ ĦĦ!ĦĦZĦĦ)¤Ħ< ! ĦG¤!Ħ Ħ GĦ*Ħ!ĦĦ 7 4ā K "Ħ ľ )ĦĦ"Ħ Ħ*!Ħ @ĦGĦĦ ľ Ħāā Ħ4ā*Ħ*Ħā Ħ āĦ*
4 EĦ*Ħ)*ĦĦĦ XĨ
Mv
Ï@" @" ā
*Ħ Ħ C2EG7HI1@IQ Ħ Ħ *Ħ *Ħ *ÿĦ "Ħ "Ħ Ħ ûć!Ħ Ħ ĕĢ *!ĦĘĆïĎĦ î"Ħ )ĦĦĦ*ö*Ħ 0ā0ā+ 44āāā ĦĉăĦĊĦ GĦ*þĦ *üĈĦ &Ħ qĦĦ*ĦGĦ)Ħ Ħ ĦĦ"ĦĦ ĄĦĦGĦĦĦ ØĦ @ā
2;,#@>B)Y #E/)8#E2Y >&2)EHY 3=.$A?C*Y$F0*:$F3'$7Y7I:A3$(Y Y yssä & #& & $& $#"&
1Ĩ Ĩ Ĩ .Q Ĩ Ĩ Ĩ Ĩ Ĩ ā Ĩ Ĩ Ĩ Ĩ ā,Ĩā Ĩ ā,Ĩ
#)Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ Ĩ +Ĩ #Ĩ Ĩ Ĩ Ĩ .Q Ĩ òĨ ĨĨ Ĩ !Ĩ ā#)Ĩ .Q Ĩ Ĩ!Ĩ ā Ĩ¨Ĩ Ĩ!Ĩ ā «
6EGIQ GB:KI6B@Q B)Ĩ Ĩ"Ĩ82 Ĩ 5Ĩ1+ +ĨJĨĨ Ĩ O ĨĨ G@Ĩ Ĩ Ĩ GOĨ WĨGÂĨ J!Ĩ gOGĨ WĨNGOĨ /Ĩ2+Ĩ #Ĩ O& Ĩ Ĩ !\Ĩ BĨ NGOĨWĨ2Ĩ ľ "ĨĨgG&ĨĨĨĨ O ĨĨG Ĩ&+Ĩ
Ĩ Ĩ Ĩ Ĩ NO @Ĩ Ĩ gO &Ĩ,ĨN& @Ĩ /ĨNê&Ĩ,ĨN& @Ĩ Ĩ NäÙ&ħ NGOĨ,2 Ĩ Ĩ /Ĩ" 3 2Ĩ,2 Ĩ Ĩ /Ĩ"Ĩ /ĨN @+Ĩ ¢Ĩ @ OĨ Ĩ Ĩ Ĩ TĚóŴ Ĩ ãĥùÛ+Ĩ #Ĩ G@Ĩ/ĨGOĨ,ĨO@ĨWG,Ĩ@ +Ĩ ăľ
ä
ä
)Ĩ <Ĩ ĨC ĨĨĨĨsĨ 5Ĩ ÞGĵŴ ŤťŴ Ô½ŴēĕĔćľ ¡ĨeĨĨ Ĩ +Ĩ Ĩ s&Ĩ Ĩ ¡Ĩ Ĩ <Ĩ Ĩ eĨ Ĩ C+Ĩ #wĨ ÖŴ ěOŴ CDIJľ ıľ úÅŢšMŐŁČŴ :Ĩ ¥ Ĩ Ĩ Ĩ ¥&Ĩ ¨Ĩ Ns +Ĩ @GsĨ ¹ e^sĨ Ĩ ŌFĊŚŨũŪşŴ ŠĸŕăIŴ ¦wĎĨ w Ĩ Ĩ Ĩ Ĩ +Ĩ ¢Ĩ @& /Ĩ^&´Ĩ e& Ĩ G&Ĩ íwðĨ à
ÿæŴ ÆâÜŴ 5wõĨ àeĨ Ĩ Ĩ Ĩ Ĩ Ĩ &Ĩ Ĩ g<Ú Ĩ N^Ĩ&Ĩ /ĨNÞáÃĨ öÃŞÄŧŴ ĀÝāxØŴ Ĝ+Ŵ G+ĮŅDIņ_Ŵ Ĩ^Ĩ W^&+Ĩ5ę!Ĩ eĨWĨ&e+Ĩ#ĨĨ Ĩ ĄOĝĞüŴ )1-B@ .Q GB:KI6B@Q
34v
*56E.QGB:KI6B@Q
aĨĨ Ĩ"Ĩ,Ĩ2Ĩ WĨĨ 23 "äĨ", Ĩ Ĩ7ŴŇğMŴ Uŀ
ľ
_Ĩ Ĩ U) Ĩ @GĨ Ĩ "Ĩ/ĨNĨ 2ĨĨ g \Ĩ ÷F)Ľ)ēĭľ)Ŵ ¤éŴ ×]Ŧ^ÇŴ ËĦĨ Ĩ gĐøĨĨ H¥$ "+Ĩ#)ĨgG&Ĩ/Ĩ2V"Ĩ N &@ĨWĨ"3 řŴ áĒŴ ĉ +Ŵ ōĘŴ DįLLİħŴ ñĀĝ 0I
"ā
ĨQ Ĩ!Q Q Ĩ Q%Q ĨĨ)Ĩ č Ĩ Ĩ !Q % ! %Q
Xā
Ĩ 23 "=Ĩ " V 2=Ĩ Ĩ I ¸ ā 2Ĩ Ĩ"Ĩ Ĩ"ĨĨ2Ĩ Ĩ2ĨĨ"Ĩ Ĩ"Ĩā Ĩ"Ĩ{ Ĩ2Ĩ Ĩ Ĩ Ĩ"ĨĨ2Ĩ Ĩ"ĨĨ2Ĩ /ÉĨ /Ĩ
`Ĩ Ĩ Ĩ Ĩ *Q Ĩ Ĩ H Ĩ ĨğĨ Ĩ Q Ĩ 4Ŵ #Ĩ "*'Q / # *Q #)Ĩ ;<' *Q ĨĨ Ĩ =>' *Q Ĩ W Ĩ +KY 5Ĩ' * Q /Ĩ'(Q Ĩ WY /Ĩ+LY ĨĨ Ĩ -Ĩ '(Q /Ĩ'Q ľ ' ®
ľ ü`$ľ ľ $ľ /Ĩ$aľ #Ĩÿ<ľ Kľ /Ĩľ Ĩ #ĨĨĔĨ Ĩ ľ ĜĞīľ đZĨ ĨĨĨ *Q(Q Ĩ #*Q Q
XY4Ŵ
ìĨ Ĩ Ĩ
5Ĩ £) Ĩ 5 Ĩ 5Ĩ é ) Ĩ Ĩ +ā )Ĩĉ)Ĩ Ĩ ĨĨ )Ĩ Ĩ Ĩ ĖŴ Ĩ 0ā +ā
93
ÐÀ
#Ĩ Ĩ ĨĨ Ĩ Ĩ Ĩ )ÁĨ ß Ĩ !Ĩ Ĩ Ĩ XĨ Ĩ ĨĨ ĕĨ XĨ)ĨĨ Ĩ!Ĩ XÏĨ ĨĨ²Ĩ+++ĨĨ+++ĨĨ LMNOľ ûľ
_Ĩ Ĩ ĨĨ ĨĨĨ XĨĨ ĨĠ$Ĩ_Ĩ ĨĨ Ĩ ĨĨ ĨĨ Ĩ$Ĩ #)ĨĨHĨ Ĩ !Ĩ QQ )Ĩ Ĩ Q ĨĨ$$+ĨĨ$$$ĨĨĨ Ĩ !ĨX³Ĩ ĨXĨ Ĩ Ĩ+Ĩ
ā
'."305 ; 7$ ,7'); 0' 79; Udm`Ypoq\}RYtb\kYtdZYi}SiwkpeY[} URS} 2,,8} Sp\m} U\Ztdom} Tovm[} /} }+a+a :Ħ Ħ¬ Ħ %Ħ
g G Ħ
PkpoqtYmt@}
EKO7JS ''S : IM7KLMKL<8QSQGMJS3EKO7JKS #35;S IM7KL
1Ħ r!Ħ = Ħ7=++Ħ+Ħ ++Ħ /+!++Ħ Ħ=aĦ `Ħ "7Ħ+Ħa77ĦĦ+uĦĦA¦Ħ 1Ħ ¨7"+Ħ Ħ 7: aĦ IS +Ħ 7++Ħ Ħ 7ēĦ +Ħ +u7Ħ PS HS4IS4 )<%Ħ Ħ =aĦ+Ħ"+Ħ\ĦHS : I S
HS :CCPS :
1Ħ
: \ 7ĦjĦ7Ħ ĦĦ
8S$:
$: !++Ħ
ĦPSQS !: ': H7aĦ j %Ħ4 8.S
8PQS :49 8P 8Q 8Q PS4Ħ\Ħ
S 2++Ħ++Ħ S. =aĦS ĦĦ+Ħ\Ħ7ĦĦ/+/Ħ\Ħ=+ĦĦ=+1Ħ H7=aĦ +Ħ/+Ħ\Ħ ! Ħ\Ħ7Ħ+Ħ /Ħ7Ħ!7ĦS aĦ+ 7ĦS 7Ħ Ħ+Ħ Ħ 1Ħ HĦ+e+:Ħ +Ħ7/+Ħ+u+ +Ħ\Ħ7Ħ7Ħ
S
S
S
S
'Ħ
)oĦ 2++Ħ +Ħ ÐĦ +7Ħ 2 29 1 1 -9 ? Ħ\Ħ?S :ĦĦo o 1 ĦĦg¦Ħ
S
S
S
S
S
S
S S
KS \Ħ :Ħ:Ħ1 1 1 ĦĦg1Ħ r!Ħ Ħ \Ħ +Ħ 7\ Ħ
%1Ħ
S Ħ7Ħ"+7 +Ħ 7=Ħ=ĦĦ 7 #+Ħ : \Ħa7ĦĦ !Ħ (S 9 % ;oĦ : 7 +Ħ !7Ħ ++Ħ Ħ7Ħ7 7/+aĦ 7Ħ S 2+Ħ7+Ħ&S ++Ħ 7 +Ħ : 7ĦĦ : 7=+Ħ
4Ħ H7aĦ+Ħ+Ħ\Ħ +Ħ++ĦĦ4Ħ
Ħ
<1Ħ H Ħ Ħ(3ĦĦ ("ĦyĦĦ Ħ ĦĦ Ħ3Ħ ĦĦĦ %Ħ AĦ HĦ ĦG ĦĦĦ*ĦĦĦ(3Ħ#Ħz*Ħ3Ħ Ħ(ĦĦ Ħ#Ħ Ħ*@Ħ2 ĦĦe Ħ%ĦG#Ħ(3Ħ*ĦĦ04@@@ĦĦfĦ ! Ħz*Ħ %@ĦH Ħ Ħ(3Ħ ĦG#Ħ(3ĦĦ m Ħ@ ĦĠĦ ! Ħ z*Ħ 9 %@Ħ
g1Ħ D(ĦjĦ ĦĦ( Ħ Ħ #&: # & 9 #&4#& ĦĦ 89 H} MĦ Ħ Î@Ħ @ĦĦ*Ħ.Ħ'ĦĦĦĦ #Ħ ĦĦ: ! Ħ (Ħ<MĦ 2 Ħ (#Ħ3 Ħ Ħ*.ĦĦ ĦĦ.'ĦĦQĦ Ħ Ħ eĦ Ħ*'Ħ Ħ
@ĦĦ Ħ "#(ĦĦJQĦ J
MĦ
&:
4 [Ħ Ħ #Ħ*.'²Ħ *.Ħ *'Ħ 9 Ħ 4Ħ ĦCĦ Ħ Ħ Ħ Ħ*.Ħ Ħ *'Ħ "# ^Ħ IĦ QĦ 3Ħ Ħ ĦĦ4CĦ ĦĦ Ħ eĦ Ħ.QĦ Ħ'QĦ Ħ Ħ Ħ*'Ħ Ħ*.ĦĦ #: Ħ ": " @ĦH Ħ Ħ"(ĦĦ &4 &
9
Ħ
MĦ [Ħ Ħ #Ħ *.'Ħ *.Ħ$ĦĦ.'Ħ$Ħ%Ħ*'Ħ$Ħ%Ħ TĦ Ħ Ħ Ħ Ħ .'Ħ Ħ4Ħ Ħ Ħ Ħ Ħ.'Ħ( Ħ Ħ*4Ħ 3 Ħ «.*'Ħ IĦQĦ 3Ħ ĦĦĦ Ħ (ĦĦ.ĦĦ*4Ħ Ħ Ħ ĦĦQ TMĦ 9
:
[Ħ Ê sĦ Ħ *.'Ħ ?*ĦBĦ ; Ħ ĉ'Ħ$Ħ ĘĦ : Ħ Ħ Ħ Ħ *.Ħ Ħ 4Ħ Ħ Ħ 7 ÊĦĦJ. Ħ D(Ħ Ħ #Ħ ĦĦ : ! Ħ (ĦJ4Ħ( Ħ Ħ Ħ : %59 * : sÉssĦ Ħ ğĦ Ħ «.'Ħ4ĦĦ ^Ħ '9
MĦ [Ħ ö*.':Ħ 4Ħ ĦCĦ Ħ Ħ Ħ Ħ Ħ Ħ *.Ħ Ħ*'Ħ "# Ħ '4Ħ Ħ .CĦ Ħ ĦQĦ ! Ħ «.E'ĦX ęĦ D(Ħ¿4$ĦAĦ Ħ 'CĦ$ĦA^Ħ H Ħ.'Ħ 9
9
RMĦ IĦ+:$Ħm ĦĦĦ ĦĦ
<1Ħ H Ħ Ħ (3Ħ Ħ#Ħ#(ĦĦ Ħ(Ħ -Ħ4
79
Ħ Ħ $Ħ %Ħ} @Ħ %Ħ 9-Ħ
AĦ H Ħ Ħ #ĦĦ ( Ħ Ħ ĦĦ "ĦĦ5`ĦL 5 ĦĦ#ĦĦ $ 9 S1Ħ ¨"Ħ !Ħ Ħ *BĦ mĦĦĦ^ ^ ĦĦRfĦ Ħ.Ħ$Ħ0 ĦfĦ Ħ Ħ (3Ħ Ħ Ħ 8S 9 *Ħ .Ħ! Ħ Ħ3Ħ ĦĦ ĦĦĦ !Ħ#ĦĦ*^Ħ
@ĦIĦ5_ Ħ5-Ħ@@@Ħ3 ĦĦ( Ħ Ħ Ħ Ħ Ħ5ðĦOĦĦ Ħ5Y$Ħ 5YĐ_6Ħ Ħ##Ħ @ĦH Ħ Ħ(Ħ ( Ħ Ħ5YĦ 9 ¶¶1Ħ @ĦIĦ QĦ 3Ħ Ħ G Ħ # Ħ 3 Ħ Ħ Ħ #MĦ H Ħ Ħ (3Ħ Ħ #Ħ ! Ħ" ĦĦ Ħ " ĦĦQĦ( Ħ Ħ Ħ !Ħ" ĦĦ Ħ#Ħ Ħ Ħ 3 Ħ Ħ #Ħ Ħ Ħ " ĦĦQ@Ħ 1Ħ 8Ħ Ħ Ħ Ħ Ħ Ħ Ħ Ħ Ħ Ħ Ħ "3Ħ 3 Ħ Ħ 3(Ħ Ħ "3Ħ 3 Ħ Ħ Ħ "3Ħ 3 Ħ 1Ħ r Ħ Ħ Ħ Ħ Ħ Ħ ĦĦ#Ħ Ħ ! Ħ Ħ Ħ %Ħ Ħ Ħ Ħ Ħ @Ħ H Ħ Ħ (3Ħ Ħ #Ħ Ħ 3!Ħ Ħ Ħ Ħ #("1Ħ @Ħ2 Ħ 3 Ħ Ħ Ħ 83Ħ Ħ ĦĦ¬ ĦA<S@Ħ Ħ !Ħ Ħ ¬ Ħ 3 Ħ Ħ č Ħ 3 Ħ Ħ } Ħ 3 Ħ Ħ 2 ( Ħ 3 Ħ Ħ H Ħ 3 Ħ RĦ D( Ħ 3 Ħ %Ħ Ħ D( Ħ 3 Ħ <Ħ ! Ħ Ħ Ħ Ħ !Ħ !Ħ 8#3Ħ Ħ 3¦Ħ ¨"Ħ (Ħ !Ħ ĦĦĦĦ ĦĦ <@Ħ 1Ħ H Ħ Ħ(3ĦĦ
"#(Ħ (: Ħ 69 AĦ 69 Ħ 69 Ħ 69 <} ! Ħ $@Ħ Ħ
ôĦ
'."305 ; 7$ ,7'*; 0' 79; &-!2/4; 6#+6&(; (8+2&; ; 3--9} Sp\n}U\Ztdon} Tovn[} /}Uoivtdonr}
/)} 8WĦg)Ħ ĦĦ Ħ3Ħ!ĦĦ # # ###& ! Ħ Ħ "&$ Ħ :Ħ:Ħ:Ħ1 1 1 Ħ:ĦgĦ! Ħ# &>Ħ# &>Ħ &>Ħ&>Ħ#& OĦ A1Ħ 2 Ħ(3Ħ ĦG"ĦĦ (Ħ Ħ Ħ3"Ħ(Ħ Ħ g)1Ħ DĦ Ħ Ħ g)Ħ( Ħ@Ħ
2(} 8qĦ Ħ PĦ>ĦdĦ
P- > Ħ )P > d > )<%Ħ )P > dĦ Ħ Ħ PdĦOĦ $Ħ >ĦĦ%Ħ1Ħ gĦ %Ħ Ħ
1Ħ DĦ P9 >ĦĦĦ Ħ(Ħ Ħ Ħ 6: >ĦdĦ 9 Ħ ĦĦ Ħ(Ħ Ħ MĦ 8Ħ PĦ>ĦdĦOĦ Ħ>Ħ g1Ħ D P O Ħ):Ħ:Ħ:Ħo ^ o1Ħ [ĦPĦ$Ħ):Ħ PĦ>ĦdĦOĦ Ħ ĦPdĦ 1Ħ [ĦPĦOĦ :Ħ PĦ>ĦdĦOĦ :Ħ PdĦ g:Ħ dĦOĦ : 2 (Ħ PĦ>ĦdĦOĦ oĦ
4(} 8£Ħ < Ħ Ħ #!$ $ #&>Ħ:Ħ Ħ!Ħ "Ħ !$#&>ĦĦOĦ $# # $&: 1Ħ #$&0ĦĦ$ (3 Ħ !Ħ "Ħ ĦOĦ !# $& >Ħ$ #& Ħ t9Ħ>Ħ$& OĦ tZĦ>Ħ# & IĦ $& OĦ oĦ 2 Ħ t9ĦOĦ jyĦ>Ħ91Ħ D(3(ĦĦ Ħ"Ħ Ħ Ħ(Ħ #& $&$Ħ :Ħ !ĦĦ
tyĦ>ĦĦOĦ tytyĦ6 tyĦ>Ħ:Ħ Ħ tyĦ6 -Ħ 2 (Ħ tyĦ
Ħ Ħ t %Ħ>Ħ&
1Ħ
< 1Ħ
5%} 8WĦ%AĦ IĦ $.9 Ħ Ħ (3Ħ Ħ ! Ħ Ħ Ħ fsa} Ħ Ħ *^Ħ }Ħ "Ħ & 1Ħ 8Ħ Ħ Ħ Ħ Ħ Ħ Ħ 3Ħ Ħ Ħ ! Ħ ĦĦ Ħ 3Ħ @ĦDĦĦ(3Ħ Ħ! Ħ Ħ 3Ħ Ħ 3Ħ ĦĦ Ħfsa} ĦĦ ~Ħ 2 Ħ (3ĦĦ! Ħ ĦĦ Ħ f}>Ħ gsa} Ħ *Ħ "Ħ Ħ 3ĦĦ đ~Ħ>Ħ 1Ħ DĦ $.19 ~Ħ $.9 2 (Ħ $ 9$Ħ :Ħ $9$Ħ:Ħ $9 :Ħ
&& & OĦ %A1Ħ
7+} 8qĦĦ §( Ħ Ħ(Ħ Ħ Ħ ^Ħ 2 Ħ ĦĦ Ħ Ħ Ħ 29 239 1 1 ^ ĦĦ z-Ħ ĦĦ ĦĦ zñĦ Ħ Ħ ĦĦ 2 9 DĦ Ħ!ĦĦ 1÷Ħ OĦ 1Ħ Ħ
8%} 8WĦĦ *4Ħ/ ĦĦĦ?*Ħ ĦÀ'Ħ/ ĦĦĦ?'Ħ !Ħ?.'4ĦX?¿*4BĦ ?*|1Ħ ?*4'ĦB ?.1Ħ r Ħ?Ï'4ĦBĦ?*0?'|Ħ Ħ?4À'ĦXĦA 6?.6?* 0?'|Ħ$Ħ ?* 0Ħ?'|Ħ 2Ħ 4Ħ$Ħ'41Ħ 2Ħ Ħ '4Ħ / Ħ Ħ Ħ &Ħ Ħ Ħ Ħ *Ħ &Ħ Ħ : r Ħ Ħ / Ħ Ħ Ħ &Ħ % Ħ Ħ Ħ Ħ Ħ #Ħ : 2Ħ 4ĦX'4BĦ % Ħ·Ħ$Ħ Ħ
:&} 8qĦ R Ħ &Ħ1;Ħ Ħ:Ħ ĦĦĦ!Ħ&Ħ1;Ħ " Ħ %Ħ Ħ jĦ1;XĦė %Ħ|/S Ħ ©Ħ 0ĦÇ %Ħ|1Ħ ;©Ħ 0ĦÇ %Ħ·1 Ħ ;©Ħ 0Ħ1 Ħ Ħ Ħ Ħ Ħ Ħ Ħ Ħ Ħ " Ħ Ħ 2Ħ /Ħ &Ħ "Ħ yĦ Ħ Ħ Ħ Ħ Ħ /Ħ ĦĦ Ħ %Ħ Ħ Ħ ®Ħ !Ħ &Ħ ĦĦ " Ħ % nĦ ! ĦĦmjĦjRfĦ$ĦjRĦ$Ħ Ħ0ĦA Ħ0Ħ%Ħ0ĦXĦ Ħ R 1Ħ
<&} 8WĦARRĦ 2ĦĦm5ĦÅĦÆf äåæĦ m65Ħ6ÅĦ6ÆfĦĦĦ/Ĕ Ħ&ĦĦĦ&ĦG#Ħ/Ħ Ħ mĦ M M Ħ²ĦfĦ !Ħ z*Ħ 9 %Ħ Ħ Ħ Ħ Ħ GĦ /Ħ &Ħ mĦ1 1 1 ĦĦfĦ !Ħ z*Ħ 9 %1Ħ 2Ħ /Ħ Ħ ÝĦ/Ħ Ħ mnĦ ĦĦfĦ Ħ -;X Ħ <<Ħ 2Ħ Ħ /ĦĦ GĦ/ĦĦ mĦ ĦĦfĦ !Ħ z*Ħ 9 %ĦĦ :
:
1} Ď<<Ħ6 /Ħ&Ħ Þ#Ħ/ĦĦ mĦ ĦĦfĦ !òĦz*ĦXĦ%ďĦ$ĦARRĦ
?%} 8¼Ħ Ħ ĦĦ9¡Ħ 0Ħ9v¡ĦXĦ9-Ħ0Ħ9ß-96Ħ0Ħ9àõĦBĦ9Ħ0Ħ9vĦ-6 9-Ħ0Ħ9v--L Ħ$Ħ 9Ħ0Ħ9v- 6 9Ħ0Ħ9v- 6 -á Ħ$Ħj9Ħ0Ħ9vĦ 2Ħ Ħ ĦBĦ 9Ħ09vĀĦ !Ħ "ĦjĦXĦ-6 -L -â Ħ 2Ħ jĦBĦ-ã -6 -6 ĦBĦĦ
/,&} 8qĦ
:
IĦÃĦ/ĦĦ&ĦĦĦ ĦĦ : Ħ.'Ħ D Ħ ?.*'ĦBĦ ?'JÃĦ !Ħ"Ħ ?*
JĦ$Ħ?J'Q1Ħ 2Ħ»þJQĦĦĦĦ»
J'1Ħ 2&ĦJ'|J
ĦXĦJQ|J'ĦĦ Ħ 5:QĦĦ J
Ħ$J Ħ '-Ħ$Ħ<-Ħ$Ħ :
RĦ
//)} 8WĦĦ IĦ4EĦ$9 Ħ ĦEĦCĦBĦZĦ Ħ.'Ħ$5Ħ 8ĦwxĦ4EĦĦ#ĦĦwxĦ.'nĦ!Ħ "Ħ4xĦNĦ.xĦ$Ħ 95Ħ 2 (Ħ .x6N NĦ.xĦ$9 Ħ 5oĦ D#"ĦĦ.x:Ħ !Ħ "Ħ NĦ.xĦ$Ħ56 95Ħ D Ħ Ħ Ħ Ħ Ħ wECĦ Ħ Ħ Ħ w.':Ħ !Ħ 3Ħ N'Ħ$Ħ 5L Z5Ħ §( Ħ Ħ Ħ BĦ5L 9Ħ }ZN5ĦBĦ5L 5N5BĦĦ } .xĦ 'Ħ
/2)} 8WĦ%Ħ e Ħ.EĦ Ħ*'ĦĦCĦ 2 Ħ *.CĦ Ħ Ħ #Ħ Ħ ! Ħ *.Ħ$Ħ*CĦ Ħ .EĦ$ĦEC Ħ Ħ 8ĦEĦ Ħ TĦĦ Ħ Ħ Ħ .CĦ Ħ .'Ħ "# Ħ !Ħ "ĦE T Ħ Ħ##ĦĦC'Ħ ĦE T$ĦC'NĦBĦ%Ħ6 N ĦBĦ%Ħ
/4)} 8WĦ)Ħ ăĦ4iĦ Ħ4'Ħ IĦ TĦ3Ħ ĦĦĦ Ħ (#ĦĦ4ĦĦ*'Ħ 2 ĦJi:Ħ 4 T Ħ Ħ.'ĦĦĦ###@ĦD Ħ4 Ħ Ħ ĦĦJ.Ħ TĦĦ Ħ ĦĦi'Ħ 2 (Ħ4i'ĦĦĦ #ĦĦ ! Ħ4iĦ$4'Ħ Ħ ?i4 TĦ$ĦĆ T4'1Ħ 8Ħ ?iJ*Ħ$Ħ< ;Ħ Ħ wîi4Ħ Ħ Ħ !Ħ "Ħ ?Ji4ĦBĦ )ġĦ 2 (Ħ ?.'4Ħ$Ħ ? T4'ĦBĦ ?i4 TĦ$Ħ?Ji4ĦBĦ) pĦ
/5)} 8qĦ %Ħ D Ħ4Ħ ĦCĦ Ħ Ħ ĦĦ Ħ Ħ*.Ħ Ħ*'Ħ " Ħ !Ħ "Ħ4CĦ Ħ ###Ħ Ħ .'Ħ Ħ wE4CĦ Ħ Ħ Ħ wE'.Ħ ! Ħ E4ĦWĦE'$ĦECĦ qĦ E.Ħ$ 4CĦWĦ '.Ħ$Ħ WpĦ IĦECĦ$Ħ9Ħ ĦE4ĦBĦZĦ 2 ĦE.ĦBĦ 9Ħ ĦE'$ĦZĦ ü Ħ Ħ Ħ Ħ Ħ wE.4Ħ Ħ wE'CĦ !Ħ Ħ 9-Ħ } Z-ĦBĦ AS-Ħ Ħ Z-Ħ } 9-ĦBĦ SA-pĦ 8 Ħ Ħ !Ħ (Ħ !Ħ "Ħ 9-Ħ4 Z-ĦBĦ AS-Ħ } SA-N)Ħ 2 (Ħ .'-Ħ $Ħ %Ħ 9-Ħ}Z-ĦBĦ AS-Ħ}SA-N)Ħ$Ħ %Ħ 2 Ħ.'ĦBĦ
/7)} 8WĦ%))%Ħ 2 Ħ!ĦĦ <Ħ <Ħ4 <Ħ ĦĦ^Ħ <ĦBĦ _¡Ħ$%Ħ ))%Ħ Ħ } Ħ )Ħ } } <Ħ
%Ħ
/8&} 8¢Ħ <)Ħ U `);Ħ0Ħ`);VĦ U Ħ);Ħ 0Ħ );VĦ $Ħ U ); L ¹Ħ );ĦĦ);Ħ0Ħ -);Ħ-); L Ħ);Ħ¹);Ħ0Ħ);VĦ hĦ U );Ħ0ĦĦ); 6 Ħ );ĦĦ);Ħ0Ħ -Ħ );Ħ-Ħ);VĦ 9 UĦUĦ -Ħ);Ħ >Ħ -Ħ);V -Ħ6 -Ħ);Ħ-Ħ);Ħ6 Ħ);ĦĦ );VĦ ĥ Ĥ $Ħ UĦL U Ħ ;V-Ħ6 U Ħ ;VVĦ Ħ Ħ $Ħ U6N%6NVĦ $Ħ< )Ħ
/:&} 8¢Ħ Ħ § Ħ ĦuĦZĦ0ĦNZĦX5Ħ ! Ħ5¾Ħ Ħ [Ħ Ħ/Ħ ĦĦ Z-65ZĦ0ĦĦ 9 Ħ Â/"Ħ ĦĦ Ħ!Ħ"ĦĦqĦ ZĦ 9
5ĢĦ Ħ
Ħ ¾Ħ ĦĦ
2 Ħ ĦuĦ 9-Ħ>Ħ WX} hĦ %Ħ0Ħ %Ħ 9-Ħ
ĦĦĦĦU : ĦÌĄºĦÌNąºVĦ
/<&} 8¢Ħ Ħ Ħ Ħ 5`65$Ħ5UĦ56VU 5Ħ0ĦVU 5-Ħ0ĦVĦ ĦĦ Ħ Ħ Ħ $S5`65Ħ Ħ Ħ $S5`65Ħ }Ħ Ħ !Ħ Ħ )Ħ $S5`L 5Ħ/ Ħ Ħ Ħ "Ħ qĦ 5ûĦc} U Ħ )VĦ c}$Ħ ĦĦĦĦMĦ 2 Ħ Ħ N}5`³ 5Ħ } Ħ5ĦhĦ Ħ !Ħ "Ħ5`Ħ6 5ĦXĦ Ħ 2 Ħ Ħ eĦ Ħ MĦ
/?'} 8WĦ< )Ħ 2 Ħ Ħ _`Ħ Ħ Ħ I} Ħ .@Ħ 2 Ħ Ħ ĦĦ Ħ _} qĦ I} .Ħ ! Ħ _ c} 9 Ħ Ħ #Ħ c} !: IB} Ħ Ħ Ħ)Ħ Ħ _} qĦ I} .Ħ ! Ħ _ c}$Ħ Ħ Ħ ##Ħ c} !: IVxfy} Ħ _ f} Ħ XĦ Ħ Ħ)Ħ 2 Ħ Ħ!ĦĦ <)MĦ
<Ħ
2,&} 8WĦ Ħ ĊĦĦ&Ħ5Y$Ħ 5Y×ĕĦ Ø Ħ ]Ħ"Ħ
5ĦYĦ {}$Ħ Ħ 5ĦYĦ´_Ħ | SĦ SĦ &Ħ#Ħ Ħ 2Ħ 5YĦ
Ħ ´ AĦ $Ħ ĦYv_Ħ 5ïĦ µ >} $Ħ ĦY _Ħ
=}
=}
&ĦĦ øùĦú Ħ 2&Ħ
:/"ĦĦ&Ħ CDE} Ħ
2Ħ
5__Ħ 9 _Ħ 9 5_-ÍĦ
ċĦĦ ] Ħ Ħ]ĦĦ pĦ
2/%} 8¼Ħ {S Ħ ÁsěĦ *Ħ /ĦĦ"eĦ&Ħ Q^Ħ HĦ ]Ħ Ħ ĦĦ/Ħ&Ħ ĦĦ"Ħ *ĦĦ #9 ÓģeĦ 8&ĦĦ] Ħ*Ħ ĦĦ "Ħ" Ħ&ĦQĦĦ Ħ Ħ&Ħ*Ħ ]ĦĦ s&ÈĦ ]Ħ Ħ" Ħ&Ħ] Ħ]Ħ Ħ Ħ]Ħ" ĦĦ ĦĦ] ĦĦ Ħ ĞÉĦ*ÔĦ Ħ Ħ #Ħ Ħ/Ħ & pĦ 2ĦĦ ;$Ħ : ; ] ĦĦ ĦĦ rs ĦĦĦ Ħ : ; ĦhĦ S Ħ Ħ #çĦ
22*} 8£Ħ A%Ħ ÁsÈĦ ): x}79 !: .: N} Ħ A} 79 A} z Ħ *Ħ hĦ x}79 !: (: N} 79 Ħ "/Ħ / Ħ z Ħ .Ħ$Ħx}79 !: (: N} 9 ĚĦ ®"/Ħ / Ħ z}S$Ħx}79 !: (: $S 79 Ħ "/Ħ/ Ħ z Ħ ČĦu Ħ]ĦĦ
Ħ Ħ Ħ Ħ Ħ Ħ $ĦA%pĦ µ } ³ 6 6 Ħ Ħ } Ħ Ħ } Ħ Ħ }
Jh N K N}}% $ $Ħ S
24%} 8WĦ)Ħ ÂĦ& Ħ Ħ Ħ Ħ G)G %Ħ } pĦ }Ħ #Ħ Ħ Ħ Ħ /Ħ &Ħ Ħ4Ħ&Ħ )G)GA<SĦĦ {G)G %pĦ 2Ħ/Ħ&Ħ#Ħ Ħ/]Ħ A<SĦ Ħ )Ħ Ħ
)Ħ
} AĦ
hĦ Ħ
:73;L 00? C7L '$ $*?D.$:$L :$L $* 0.1? J%:?L .F@L
%+? J$:>L 7*L I.0!.L '? :%L 1$8L
1*>'+*' >2(++'+*'$0 )' %0'''?
04!&L )+),)? 1$H$?L L :&204#%:L 7+L *?I.$4L #0H0#$#L
JL .? 1 $:CL 04>C$04L I>L
)+)+)? 7:4L 74L
:0#J L .$L 4>I$:L 0>L
4>L $1..+? 'CL 37? #$47C$L C.$L 4F<5 (:L 7*L AG!.L 6#0,0CL 04C$,$:> L 374,L C.$>$L 47? 04C$,$:@L 1$CL 57? #$47C$L C.$L 4F<5 $:L 7*LC.7?%LI.0".L$4#LI0C.L$? JL >J22$C:JL C.$L 4F=4 $:L 7*LC.7>$LI.0!.L
$4#L I0C.L '? 7:L )? 0?L 1?7L $9F1L E7L 57? $4!$L %37?
?
'57?
?
%4#L I0C.L ? 7:L
$4#L I0C.L %'? 7:L
%47&?
%-7?
?
%?
$4#L I0C.L '%? 7:L )%?
)4#L I0C.L $? 7:L $? /FAL 47? $48 ?
$L .H$L 3?
)? 3&?
)? >?*
'57?
+47&??+576?
+47&??$47 $47&? !
)47? ?%49&?
.? JL 0C$:C04,L I$L ,&CL 3/?
$1..+?
4@L $.)? F@04-KL C.$L *!CL C.CL
' ?
=? ?
L
?
':? ?
?
?
' 6L ? 6
6L ?
?
"#! L I$L .H$L
' 6L ?
' 6L ?
6L
' 6L
) 6L ?
?
?
';?
?
6L
.FBL I.%4L %?
$'? '0? )? )'? ).?
)1?
%.)?
?
?
?
?
'
'."305 ; 7$ ,7'); 0' 79; Uem`Ypoq]}RYtb]kYteZYj}SiwkpeY[} URS} 2,,8} Sp]m} U]Zteom} Up]ZeYj} Tovm[} D Ħ ĦĂ Ħ %Ħ
S Ù Ħ
PkpoqtYmt@}
LL7DHLS3KSD3EQS IM7KL
:
!Ħ Ħ Ħ Ħ &Ħ Ħ Ħ Ħ Ħ &Ħ Ħ #Ħ &Ħ /Ħ } ĦĦ"ĦĦĦ&!Ħ !Ħ&ĦĦ ĦĦ % )S ĦAĦĦ" " " y} seĦĦ"ĦĦĦĦWĦ Ħ G}
Ħ Ħ Ħ Ħ Ħ Ħ 0} : 6} : ;} : u} : Ħ : Ħ : Ħ : %
èĦ ý Ħ Ħ u ĦH H S Ħ &Ħ Ħ Ħ Ħ &Ħ Ħ<S nĦ Ħ/=S %02:6ĦĦ/=S 9 / =BS4Ħ 8Ħe#ĦĦĦu Ħ Ħ)ĦĦĦ<Ħ D!Ħ Ħ Ä Ħ Ħu Ħ ĦĦĦ/Ħ&ĦpĦ
:
IsĜĦ ,Ħ/ĦĦ"ĦĦ IĦ 1 S1 S 1@S /ĦĦ Ħ&Ħ,GĦ/Ħ &Ħ %nĦnĦnĦĦ 1 ĦÑ,Ħ6 nĦ,y} Ħ Ħ 8/: 3: 81: Ħ Ħ Ħ ,Ħ #Ħ &Ħ #Ħ Ħ &c} 9 >S &f"} D!Ħ Ħ
4 IĦ3 4SS Ħ ,Ħ/Ħ "ĦĦ "ĦĦ ,Ħ " Ħ
4FB3 3S44 3S 0Ħ4 }
) Ħ
} ,L 4 Ħ
'."305 ; 7$ ,7'*; 0' 79; &-!2/4; 6#+6&(; (8+2&; ; 3--9} Sp\m} U\Ztdom} Up\ZeYi} Tovm[} Uoivtdomr}
4 DĦ Ħ Ħ b =Ħ F FĦ FĦ Ħ S FĦ Ħ =Ħ = Ħ S FĦ Ħ 9GeFĦ IĦ Ħ b =Ħ &Ħ S /Ħ &4 Ħ = Ħ Ħ b =Ħ &Ħ S /Ħ %4% 4 2=Ħ Ħ b =FĦ&Ħ4:Ħ C:Ħ = Ħ T Ħ &"4%% 4&& 4= Ħ %"4Fb" Ħ 9
(9
9
2FĦ 8FĦ
S$Ħ% &4} %!4$Ħ &4 %&4}%4
2&Ħ T ĦhĦ S [=Ħ Ħ G= Ħ =Ħ C T :Ħ CªĦ$ĦR T Ħ 2FĦ CªĦhĦ RS h C4Ħ 2ĦFĦ ?C4ćĦ$Ħ?Cª4Ħ 4 DFĦĦĦF==ĦFFĦĦbbFĦ&ĦĦ F=bĦF"Ħ=F:Ħ !ĦĦ 9_ :Ħ 99 !,9 !Ħ !+"9 4 Ħ@ĦD=bĦ4$ĦĈĦ
*4
,
09
ĦĦ
&Ħ !bĦ !ĦĦĦ 9 4 ĦMĦ r=bĦĦ$Ħ444 Ħ D=bĦ Ħ 4 4 4 4 F.} 4 4 4 Ħ Ħ !Ħ FĦĦ!Ħ = ĦĦFĦAĦFoĦ 4
Ħ Ë Ħ&Ħ ĦĦĦ Ħ Ħ Ħ &Ħ Ħ &Ħ%lĦ9: {Ħ [&ĦĦĦ PĦ %lĦ0ĦĦĦĦ Ħ
Ħ Ħ
Ħ &Ħ :Ħ Ħ/Ħ &Ħ Ħ &Ħ PL Ħ %lĦ0ĦĦ Ħ PĦ0ĦĦĦ /#Ħ/ Ħ Ħ r Ħ Ħ
Ħ £Ħ éĦ %lĦ0ĦĦĦ-Ħ%lĦ0Ħ{Ħ ĦnĦó%lĦ0ĦĦ Ħ
2 Ħu Ħ ĦĦĦ&ĦĦĦ 2Ħ Ħ Ħ Ħ !ĦĦĦ Ħ / Ħ HÒĦI#Ħ 2 :Ħ Ħ ËĦ?S Ħ Ħ ~ĦL Ħ Ħ%lĦ0ĦĦ 2 Ħ ~%lĦ0ĦĦ 9 Ħ Ħ %lĦ0Ħ Ħ r Ħ ~ĦĦ%lĦ0ĦĦ Ħ Ħ Ħ 8Ħ # Ħ Ħ Ħ /Ħ Ħ &Ħ Ħ Ħ %lĦ0Ħ{Ħ Ħ IĦ *Ħ 9 % ĦĦ 2 Ħ
},Ħz #} IĦ L} /Ħ Ħ & Ħ &Ħ /Ħ Ħ *Ħ ! Ħ ,Ħ0ĦGĦ
,Ħ
NLO} hĦ , 0Ħ
Ħ Ħ Ħ,Ħ0ĦÚĦĦ (/: ĦĦĦ/ĦĦL$} D Ħ (/4:(1: ĦĦ Ħ,ĦĦĦ*Ħ Ħ Ħ,Ħ0ĦG#ĦĦ (/: Ħ ĒĦ&Ħ Ħ,Ħ0Ħ G#Ħ Ħ (1: &Ħ##Ħ Ħ¤ĦcĦ¥Ħ>S½Ħ? S 2 Ħ ~Ħ
QLQ} ; ëĦ
4/9
r Ħ
,,Ħ0Ħ ?S ,,Ħ0 ĦĦ
,,Ħ0 Ħ ,,Ħ0 Ħ
¥Ħ ?S6
,ĦKĦĦ,ĦL }KĦ KĦ,Ħ ,ĦKĦ Ħ,ĦL }KĦ KĦ,Ħ ÍĦ
[Ħ Ħ/Ħ !Ħ Ħ ,L c, 0cĦ ,K, ¥ X%Ħ ,Ü c,Ħ0ĦcĦ 6 ,ĦKĦ,Ħ &Ħ#Ħ ¤Ħc½Ħ ,6 {NĦ [&Ħ,ĦĦ Ħ Ħ ,ĦKĦĦ,Ħ6 {ĦKĦ KĦ,Ħ ,ĦKĦ ,ĦL }KĦ ĦKĦ,Ħ
YÛ¯-Ħ
;
)9
,L c,Ħ0ĦcĦ Y±¯ pĦ % ,6 c,Ħ0ĦcĦ
[&Ħ,ĦĦ Ħ Ħ Y -Ħ êĦ } ,6 ¸,Ħ0Ħ¸Ħ ; ,ĦKĦ,6 }KĦ KĦ,Ħ ,6 ,N Ħ vhĦ ¤Ħ %_-%Y-Ħ 9 %Y±¯-Ħ ,ĦKĦ,6 }KĦ KĦ,Ħ ,6 ,N Ħ * ,6 c,Ħ0ĦcĦ
9
)Ħ
4 }ĦĖĦk"ĦĦ&kĦ°Ħ kĦ /4!Ħ4 F}/4 l} &Ħ/#4 Ml} Ħ/#4 4 %. +$$44 %$44%4 4$4 l % 4 [&Ħ/44 %4 ĦĦ &9
l} F} *4 G
*+4
/4 * ,4
l}
Fl} / 4
!Ħ "Ħ &9 l 4ĦĦ/#-%.,4 2Ħ !Ħ Ħ/#4 4 %.$$44%$44%4111$4 l %4 [&Ħ/4^} %4 Ħ kĦ eĦ Ħ "Ħ kĦ %4 Ħ Ħ %%4 4 Ħ/#4 Ħ Ħ/4^}%4 2Ħ %'$$44%$44%4
$4 l %4
$%$)44.}14114 4 $%44l 4 Ħ/#4
8Ħ Ħ kĦ Ħ Ħ&ĦĦ /"Ħ k Ħ Ħ Ħ k Ħ &Ħ l} "Ħ kĦ ĦĦ "/Ħ/°Ħl!} ÿĦ®Ħk&kĦ "/#Ħ/°Ħ/#4Ħ 2ĦĦ /#4 4 %'$$44%$44%4123$4 l %4 D Ħ /4 ^} )4 !Ħ "Ħ /#4%(4 9 4 Ħ Ħ /#4 4 $$44%$44%4 Ħ /#4 4 %.$$44%$44%4111$4 l %4
4
$4 l %4 Ħ
',% /.0$; 2&$+ 2'! *;*6+/' #;; # y
ApJiIEvy y>Evy y
.48%2; ; + 16%45(/.4 ; .5%2;9/62; .48%24;/.;5'%; .48%2;4'%%5;037($%$ ; /2;5'%;-6,5(0,%; #'/(#%;16%45(/.4; %.5%2; /.,9;5'%;,%55%24; ; "; #; $; /2 %; #/22%40/.$(.&; 5/;5'%;#/2: 2%#5; .48%24;(.;5'%; .48%2;4'%%5 ; /2; 5'%;/5'%2; 4'/25;16%45(/.4; 82(5%;9/62; .48%24;(.; .48%2;4'%%5; .$;4' $%;5'%; 00302( 5%;"6": ",%4;"%,/8;9/62; .48%24 ; /;45%04; 2%;.%%$%$;5/;*645);9/62; .48%24 ; #';16%45(/.;# 22(%4; + - 2+ ; /;# ,#6, 5/24; 2%; ,,/8%$;
CREoyZiyoRJyp`ZoiyIZQZoyaKy ! ' Ey $y
Fy $y
Hy$y
Iy!$y
Jy#
CREoyZiyoRJyqE\pJyaKy ;S y PS )y PS y y S y PS y PS y y S y *'y Ey $y
Hy$y
Fy $y
Iy$y
Jy
y CREoyZiyoRJy\EioyorayIZQZoiyaKy y S y S y S y S Ey$y
Fy %y
Hy$y
Iy$y
37 37 37
S y S y S 'y
Jy!
9ary^E`vy wJhaJiyIaJiyoRJy`p^FJhyy QS #y PS '/BOS Ey /2B Fy#$y
Hy $y
Iy $y
17 17 27
QS y QS y OS
yJ`IyrZoR'
Jy
:`y Ey iepEhJy ; \Joy ;FJyEy caZ`oy a`yoRJy iZIJy ;ipHRy oREoy ; + ; E`Iy ;FJy oRJ ^ZIcaZ`oyaKy ; :KyoRJyEhJEyaKyoRJyohZE`Q\Jy ;Ziy yrREoyZiyoRJyEhJEyaKyohZE`Q\Jy
MS
Dy *2B Fy "2B Hy "*2B Iy #2B Jy #*B
,
ARJyqD\pJiyaKy byrRTHRyiDlTiKvylRJyJepDlTa`i b$BS 0(;S %Bby bfy + b ?f
/B
+
% B
^DvyFJyKap`IyFvyia\qT`Qy
> ,B+ 2B Fy !%>$ "#"> '#,B+ 2B Hy #0>$ !!> !.'B+ 2B Iy #0>$BS !!*> !%#,B+ 2B Jy %0>$ !'> !.'B+ B Dy >$
.
BiT`QylRJyITQTliy "B#B%Bya`\vya`HJylayKag^yDy ITQTly`p^FJgyRary^D`vyaKylRJ^yDgJyITqTiTF\J Fvy ""5 Dy '3B Fy *2B Hy ,2B Iy .3B Jy /
/
:KyDyca\vQa`yRDiyTliyip^yaKyT`lJgTagyD`Q\Jiyi^D\\JgylRD`y #*?BrRDlyTiylRJy^DsT^p^y`p^FJg aKyiTIJiyaKylRJyca\vQa`' Dy ""2B Fy "#2B Hy !%3B Iy "'2B Jy "*
0
ARJy\Dily #BITQTliyaK
#*BS #*$BS #*&BS )*+S S #*$+B
AB7S
Dy 2B Fy *2B Hy #*2B Iy *2B Jy .*B
"
@pccaiJy yITilT`Hly`p^FJgiyDgJyHRaiJ`yKga^y "B#B lRJy\DgQJilycaiiTF\JycgaIpHlyaKylRaiJy %B`p^FJgi' D . &
<2B
F
<&2B
H
""
@pccaiJy ; B";; BD`I
"#
4T`IylRJyJsDHlyqD\pJyaKy
< <8S
I
BB%
<& .< ,2B
J
<& .. ,
8; + :; + #*B 4T`IylRJyqD\pJyaK
rRJ`y >B+ #,&#'&
y @pccaiJy > *+ + "B 4T`IylRJyqD\pJyaKy
#B
"'
@T^c\TKv
B"B 3;B"B 8B"B ; B"B 7; "B ; @" AB "
7 67 7
7 07
ARJy OQphJy ; TjyDyhJQp\DhyRJtDQa`y 3qD\pDlJylRJyepalTJ`l ,hJDyaKyRJtDQa`y ; ,hJDyaKylhTD`Q\Jy ;
",
@pccajJy 4;D`Iy 5;DhJylrayhJD\y`p^FJhjyjpHRylRDly 4;;5; "BD`Iy 4; ;5; aKy 4;;)
".
4T`IylRJyqD\pJy T`ylRJyjT^c\JjlyKah^yaKy
"/
:Ky 4;;yD`Iy 4; ; ; + yLV`IylRJyqD\pJyaKy 4; ;9
"0
ARJyITDQhD_yjRarjyDyjJQ^J`lyaKyDyHThH\JyjpHRylRDly ;TjycJhcJ`ITHp\DhyFTjJHlahyaKylRJyHRahI ; 8TqJ`ylRDly ; ",BD`Iy ; y O`IylRJyITD^JlJhyaKylRJyHThH\J
;y
",.B 4T`IylRJyqD\pJ
y
=S
;S
GS
<S
,ycD\T`Iha^Jy`p^FJhyTjylRJyjD^JyJTlRJhyhJDIyKha^y\JMlylayhTQRlyahyhTQRlylay\JKlyKahyJtD^d]J "#"BTjyDycD\T`Iha^Jy`p^FJhy 9ary^D`vy ITQTlycD\T`Iha^Jy`p^FJhjyDhJylRJhJylaQJnRJh( y =Jly (+FJyDychT^Jy`p^FJhyjpHRylRDlylRJy`Jtly\DhQJhy`p^FJhyTjyDycJhKJHlyjepDhJy 4T`IylRJyjp^y aKy D\\y jpHRy chT^Jy `p^FJhjy 4ahyJtD^c\JyTKy vapy lRT`[ylRDly ""BD`Iy "%BDhJy lrayjpHRychT^Jy `p^FJhjylRJ`ylRJyjp^yTjy y
y :`ylRJy OQphJyFJ\aryTKy ; S ;-y ;; ;; ;;1y ;; ;;7y
lSJ`y rRDly Uky=4B
y
7
=BIJQhJJk
+
!+
#%
8WqJ`ylRDly =B
7
#6)B %7$BrRJgJy 6BD`Iy GDgJycaiWlWqJyW`lJQJgiy P`IylRJy\JDilycaiiWF\JyqD\pJyaK 7
D(S
#'
:Ky =$By= #+
#*
:`ylRJyIWDQgD^y+-y 25y /.y 8WqJ`ylRDl +.-/y #* B-.y P`Iy .5
7
BP`IylRJy qD\pJy aKy =(B %=$By%
<S
#,
7
BD`Iy 2.2/y *
HS
@S
4W`IylRJyip^yaKyD\\ycaiiWF\JyqD\pJiyaKy 6BipHRylRDlylRJyKa\\arW`QyJepDlWa`yRDiygJD\ygaalyW`y =1
= 69By=$ %=By#$B B 7
#-
:Ky lray caiWlWqJyW`lJQJgiy :BD`Iy ; BFalRyFWQQJgylRD`y +iDlWiKvylRJyJepDlWa`
#*$BB:$B #'$BB;$ 7
P`IylRJyqD\pJyaNy:BB
#/
7
B
@pccaiJy 6B B7B BFS %& BD`Iy
2S
7
S
7
B 4 U`IylRJyqD\pJyaK 'B
LS !.31; 0"; !06-; .-; 3&"; '0 4,#"0"- "; .#; ; '0 +" ; 9; ).'-'-%; ++; 3&"; !.31; 3.; .-"; -.3&"0; 9; 130'%&3; +'-"1; ;3&"; ,7',4,; -4,"0; .#; 0"%'.-1; 3&3; -; "; #.0,"!; '3&"; '0 +"; '2; .4-3"!; .0; "7,/+";6&"-; LS :S ; 3&"; ,7',4,; -4,"0; .#; 0"%'.-1; '1;
+
&3; '1; 3&"; ,7',4,; -4,"0; .#; 0"%'.-1; 3&3; -; "; #.0,"!; 6&"-; KS 9S ;
.4; 6+*; ; 1/'0+'-%; ,:"; .-; 3&"; 03"1'-; /+-"; 1; #.++.61; 1303'-%; 3; ; ; -! 3&"; $013; $5"; 13./1; 0"; 3; + ;+ ;;+ ;;+ ;; -!; + ; .40 -'-3&; 13./; '1; 3; 3&"; /.'-3; ;; -!; 1.; .-; 1""; 3&"; !(%0,; "+.6 ; &3; '1; 3&" 8 ..0!'-3"; .#; 3&"; /.'-3; 6&' &; 9.4; 6.4+!; 00'5"; 3; .-; 9.40;
3&; 13./
6
0
N 5
1 -7 7 7 7 7 7 7
7 7 7 7 7 7 7
7 7 7 7 7 7 7
7 7 7 7 7 7 7
7 7 7 7 !7 !7 7 /7
"##$%7 .7
1
%+$%%%,7
;
/7
y :`ylRJyKa\\arT`Qy PQpgJy+/&y +-y
+
%+
+
%#
+
@pccaiJylRDly ES !;#;DgJyITilT`Hly`p^FJgiyipHRylRDl ! E
L` X IylRJy qD\pJyaKy
%%
3
! ## Ey&y
JI'S
6T`IylRJy`p^FJgyaKyJqJ`yITQTliyT`ylRJycgaIpHlyaKylRJylray ITQTly`p^FJgi
##########BCS """"""""""y
6T`IyD`y T`lJQJgy uylRDlyiDlTi LJY iylRJyJepDlTa` u \ u """uy S "y&yy
%*
0JlJg^T`JylRJyiJHa`Iyi^D\\JilycgT^JyKDHlagyaK yS +
BS y
#yS y #BS y
%yS y %BS '
,B
# *4S S # *BS y
HUZRJ^\`Ph AJcSPXJcUMJWh CWeX^UJNh
@dYT[_h GOLbT[Yh G[VdbT[Yah % @py8 [
/-h 1h /,h /h /hns[%"ygpW\ /,h # %ns[%"
( @py8 U I\{ Q IhK2h IhK*h1hI*hh+I)Kh-h K& hK*'h Me\p _"0 %0 1h "0(*h # % - @py; U %h(h%&$hh(""0 # gh (""0 (""2 E\pW\{e\lTy{{s[gdg{ygy%0 ( / @py< \ A
Ws}p{gpd{e\p}nU\x s^ ^TW{sx 0gp{e\vxs[}W{ \\Tygl
y\\{eT{{e\ p}nU\x 0" 0" 0" \p[y g{e # %(\xsy 0 (0 0 @py8 [
D9h
Me\{s{xgTpdl\yTx\ygnglTxTp[ ;h 1h ( E\pW\@x\TE;62h/ @x\TD9Eh3h (" E 2 @py; T % L}Uy{g{}{gpd 2h ] 5gp{s{e\ bxy{ \w}T{gsp \ d\{ T
F
/h @py; \ H^{e\axy{[gdg{gy% {e\p{e\p}nU\xgy[g~gygUl\U
%%gnvlg\y{eT{{e\-x[[gdg{gy /Tp[{e\x\Tx\{s T
y{sv}{(Tp[- Me\WTy\y e\p{e\cxy{[gdg{Tx\( - / x\yv\W{xg~\l
WTp U\[\Tl{ g{e ygnglTxl
E\pW\{e\ {s{Tlp}nU\xgy/ ( # 5 5 @py; W Me\y}ns^gp{]xgsxTpdl\ygy ( %5" # (""0 E\pW\{e\nTgn}n # gy%- 6 @py8 U Me\lTy{ ( [gdg{ys^ (""0r gy(0\W\v{ ## %egWedg~\y0 E\pW\{e\lTy{ ([gdg{y gy0h (""/ (0ns[ %"" # "0
/h
%" @py; T Dxsn{e\g[\p{g{
.SV> SV& S V& sp\y\\y{eT{ e\p SV gyb\[ {e\ vxs[}W{gy{e\mTxd\y{ e\p hS Vi gy{e\ynTmm\y{ Js m\{ S # V # Z U\ & [gy{gpW{vsyg{g~\gp{\d\xyy}We{eT{ SV Z -q Tp[{e\ vxs[}W{SVZgy{e\Ugdd\y{vsyygUm\ H^STp[V[g`\xnsx\{eTp( {e\pU
gpWx\Tygpd S U
% Tp[[\Wx\Tygpd V U
% {e\ vxs[}W{ SVZ gmm U\Wsn\ Ugdd\x E\pW\ S Tp[V [g`\x U
T{nsy{ ( LgngmTxm
V Tp[Z [g`\x U
T{nsy{ ( Kp{e\ s{e\x eTp[ g^Z>V(> S. {e\psp\WTpgpWx\Ty\{e\~Tm}\ s^ SVZ U
[\Wx\TygpdZ U
% Tp[gpWx\Tygpd S U
% Ls {egy WTy\gy x}m\[ s}{ LgpW\ SVZ gy [g~gygUm\ U
& g{ gy ps{ vsyygUm\ {eT{ Z> V%> S & sx Z V(> S & Ls\n}y{eT~\Z=V% > S( E\pW\{e\{ex\\p}nU\xyTx\q % qTp[q% Me}y {e\gx vxs[}W{ gyq %qq%=q q
%% @py; (""0 V=(""0S Tp[Z=(""0V # VZ=(""0S V E\pW\
VZ >(""0 SV
%( @py; (""0 I\{q>(""0 Me\p #&>q q &>2q*( E\pW\
fh=(""0
%- @py; % Me\\vx\yygsp> # $# $#&# h $&$h #$& L}Uy{g{}{gpd # $ y\\y{eT{{e\ \vx\yygsp= & $&&=%
% sp\
%. @py; %""-
% % >%""- Me\ \vx\yygsp> % &= "# "# #
%0 @py8 & % I\{ U\{e\W\p{\xs^{e\x\d}mTxe\Tdsp Me\p@x\T B8h:h ? @x\T 48h> .h!# @x\T 478:<>#h Me\x\y}m{ ^smmsy %2 @py; (""0! %24> #h% #$& # #$
24 &=(""0 E\pW\ #& $&= #&h$# #$&$ (
%4 @py8 / Bsnvm\{gpd{e\yw}Tx\y sp\y\\y{eT{{e\\vx\yygsp>
h
%5 @py7 %%" % @y #& # " \eR~\ #&
#
# % % % 0 Me}y #& h # #& #&h& +# %%" #
#
%6 @py8 (" I\{ !& U\{e\xR[f}y Me\p "&
! &h5& # !
#
%" Me}y{e\[fRn\{\xfy("
(" @py9 6"" L}WeRtRlfp[xsn\p}nU\xfyWsntl\{\l
[\{\xnfp\[U
f{yaxy{{ex\\[fdf{y Me\x\ Rx\6 %" %"WesfW\ys^{e\%y{ (p[Rp[+x[[fdf{y x\yt\W{f~\l
e\pW\{s{Rl # 6"" (% @py7 + H^u h% # & u #h % LfpW\ufytxfn\ # # ( E\pW\u# +fy}pfw}\
(( @py8 0." A
[f~f[fpd }t fp{s{xfRpdl\y f{e x\yt\W{ {s RW\p{\x # sp\ y\\y {eR{ {e\x\ Rx\ y\~\p{xfRpdl\y^sxn\[\d B58h B7:h \{W Me\ nf[[l\ Rpdl\R{#fyR[[\[}t {fW\ E\pW\ #& # 4 # %5" 4(" # 0." (+ @py8 %000( KUy\x~\{eR{+ F# Rp[( G V L\{V (V& Ne\p\y\\{eR{( G# Me}y{e\ynRll\y{ tsyyfUl\ ~Rl}\s^# fy2 E\pW\ #&# (2 1 %000(
(. @py8 ( Pyfpdlspd[f~fyfsp \d\{ # #&h+# #&h# # # %h( (0 @py9 2" Cxsn{e\\w}Rlf{f\y
8h
5h 7>h 78h 7>h 78h
:<8
%0"
#
8>h
7<5h
587:h 78h
(0" Rp[ \eR~\ 8>h# 2" &
(2 @py9 + # +( # # # " # %sx # ( (4 @py8 (%% LfpW\ # # h# (%%
h
#
.""6 # (%%
6
#
%6 efWe fy R txs[}W{ s^ txfn\z E\pW\
(5 @py7 ' I\||gpd = j \ d\| S= Vj V= Xj Rp[X# Sj Me}y S= Sj, Me}y j= ' Rp[
S 0h V=X Me\ x\y}l| ^sllsy (6 @py7 04 Kp\ YRp Ys}p| [gx\Y|l
|eR| |e\oRgo}op}oU\xs^ x\dgspygy04 +" @py7 + KUy\x\ |eR| |e\ y|st p}oU\xy ' 6 (0 .6 Rx\ R| |e\ls\x xgde| Ysxp\xy Me\ tsgp|" qgyR||e\y|stp}oU\x(q') q' # .q)+q Qe\pq# (( \eR\.(()+ (( 0h(""( Ls|e\tsgp|" ((gy|e\(""(|ey|st Me}y|e\ tsgp|+ (( gy|e\ (""0|ey|st +' @py7 (0 Ptsp xs|R|gpd 4=:h Rp|gYlsYkgy\ 6" |exs}de 4h \ d\| Ryw}Rx\ Me}y |e\ Rx\R gy0*=(0 +( @py7 ' KUy\x\ |eR| V XX S) .V XX S= " E\pY\ V X X S)=" Me}yV X# X S ++ @py: '" (((((((((( # '"&$ '=(((((((((("""""""""" (((((((((( #
((((((((('44444444445
+. @py7 '" Ptspx\xg|gpd |e\ \w}R|gsp Ry * '"', 666'=" sp\y\\y |eR| = gy|e\spl
gp|\dxRlysl}|gsp +0 @py7 ''
*$$1 ' ' Me\y}o= O * O' 0h .h # (""0# (""3 !# ."'' h # (""0 # (""3(""0 Ptsp #
ygotlg^
gpd \ y\\ |eR| |e\ y}o= 0 # ."' # '+.""6= 0 # '' # ."' !# '('5'6! LgpY\( + 4[s\yps|[gg[\."'Rp['('5'6 |e\y\Ysp[yoRml\y|txgo\^RY|sxgy''
'"
,?I:/OLR8] &/U<8F/U?2/C](DXFO?/5] ,&(] ,8I?LR],82U?LI
/K8H72O Q & -2OQ &
&
$FOLRU/IU +27; ; Q /524$-+2; +4; 9-5; +272; -+; 4"; +27; 2"4; .*6$ ; ; 4"; )5(4$.(; "-$; /524$-+2; +4; 9-5; +272; $+; 4"; +27; 2"4; 9; 2"$+!; 4"; 5(2; -+4$+$+!; 4"; (442; UÄ ]
] YÄ -; [ Ä -2.-+$+!; 4-; 4"; -4; +272 ;
; 4"; -4"; 2"-4; /524$-+2; $4; 9-5; +272; $+; +27; 2"4; +; 2"; 4"; ..*: .$4; 5(2; (-7; 9-5; +272 ; -; 24.2; ; +; 4-;%524$ 9; 9-5; +272 ; "; /524$-+; $2; Q )& ; -; (5(4-2; ; ((-7 ;
0<2JQ =HQ J<8Q H?2>>8HJQ DCH=J=L8Q DF=?8Q :25JCFQ C:Q J<8Q =BJ8;8FQ & Q &
U Ä &
]
&
]
&
YÄ
"]
&
&
0<2JQ =HQ J<8Q L2>K8Q C: & Q & & & Q
&
&
U Ä &
]
]
&
&
!]
&
#
.CB8Q C:Q J<9Q 23CL8Q
,8JQ .; 38Q 2Q F82>Q BK?38FQ HK6OQ CB8Q HC>KJ=CB
/<8B
U Ä .; LÄ
.;
&
.; &
YÄ
.;
[ Ä
.; &
0<2JQ =HQ J<8Q IK?Q C:Q J<8Q >2HJQ JMCQ 7=;=JHQ C:Q J<8Q =BJ8;8FQ & Q & Q & Q
U Ä &
Q
]
&
YÄ GÄ
Q &
[ Ä J
!$"#& & =HQ 8EK2>Q JCQ &
U Ä &
]
&
]
&
&
!]
&
#]
E !Ä c¢Ä Ä ~¨~Ä ; ; ~¢Ä ; ~¨Ä ¨Ä ¦¥¢¬Ä ¥¢Ä Ä ¨Ä ±¥¬Ä ¢¨Ä ¬Ä vÄ ¢¬Ä ; ~¢Ä ; ~¨Ä ¦¨¥®Ä ¥Ä Ä ~Ä ; t®¦¦¥¬Ä ~Ä ;PÄ ; ~¢Ä ;PÄ25¾Ä~¢Ä ; \¢Äį~®Ä¥Äµ!
UÄ :EÄ
&
] ?5Ä WÄ ALÄ YÄ B?Ä "] E+
lÄ µÄ ~¢Ä ·Ä Ä ¦¥¬¯Ä ¨~Ä ¢®¨¬!Ä y~Ä ¬Ä Ä ¬~¬Ä ¦¥¬¬Ä ¯~®Ä ¥Ä 2EÄ 2+LÄ µ·R ·Ä U Ä 2EÄ
] 2LÄ WÄ 5?Ä YÄ :+Ä
"] :E
5µÄ 2Ä ¬Ä¨Ä~¥®ÄÄ µ 2Ä ¢Ä·ÄPĵ!Ä y~ĬÄħ®~¥¢Ä ¥¨ÄĨ~¦Ä¥ÄÄ~Ä¥ÄĨ¥¢RÄ
L!Ä c¢ÄÄW~¨¬~¢Ä¦~¢Ä Ĩ~¦Ä¥ÄÄ ®¢¥¢Ä ·ÄPÄ 2Ä 5µÄ 2 ÄÄ 2 "] ·PÄ 5 µÄ
U Ä ·
Q
t¦Ä
·
5Ä 2 Ä Ä 2Ä Ä
]
2Ä Ä5Ä
WÄ ·ÄPÄ
5µÄ 2 2 ÄÄ
YÄ ·ÄPÄ
2 ÄÄ 5Ä
Ä ±¨ÄµÄ ¬Ä~¢¸Ä ¦¥¬¯Ä¨~Ä¢®¨#
2 7
ÄÄ YÄ µ7Ä 2
WÄ
[ Ä
2+&Ä lÄµÄ ~¢Ä ·Ä Ĩ~Ä¢®¨¬Ä¬®Ä~Ä µ7Ä Ä·7ÄQÄ5µÄ 5·ÄQ 5!Ä y~ĬÄÄ~¨¬Ä ¦¥¬¬Ä¯~®Ä¥Äµ7Ä Ä·7RÄ Ä xÄ U Ä 2+Ä ÄLw8Ä ] LÄ E [ Ä p¥¢Ä¥ÄÄ~¥¯Ä
WÄ EÄ Ä?xÄ
25Ä
YÄ AÄ 5Ä ¹
22" ]¢ÄĨ~¬Ä¢¨Ä¬¬Ä~¢Ä 5Ä Ä 25
!"Q ¬Ä ~Ä ¨~¢Ä ¬®Ä ~Ä "QQ M+¾ Ä t®¦¦¥¬Ä "QQ 2BEÄ
Ä !QQ 2EMÄ
~¢Ä Ħ¨¦¢®~¨Ä¬~¢Ä¨¥Ä "Q ¥Ä !Q ¬ÄµÄ
!Ä ]¢Äį~®Ä¥Äµ&
2: ]¢Ä Ĭ®Ä¥Ä~ÄÄ ¨~Ä ¢®¨¬ÄµÄ ~Ä ¬~¬¸Ä Ä §®~¥¢
2? v¨Ä¦¥¬¯Ä¢¨¬Ä~¨Ä¬®Ä~Ä¸Ä ¨Ä¨¥Ä~Ä¥¨Ä¸Ä ~Ä¥¬ÄE Ä cĬ ~¬¥Ä ¢¥±¢Ä ~Ä Ä ¦¨¥®Ä ¥Ä ¬Ä ¨Ä ¢¨¬Ä¬Ä5K+K#Ä y~Ä ¬Ä Ä ¬~¬ ¢¨Ä~¥¢ÄR 2BÄ!Ä GPQd¦¸Ä AQ
5++:65++?6Ä Ä5++BÄ 5++B65++?6 5++:Ä !Ä 5++?; 2Ä 5++?6 25++?;Ä Ä2Ä
2E W¥¢¬¨ÄÄ ®¢¥¢ ³Ä
2Ä
Y[] ]¢Äį~®Ä ¥
:²Ä Ä :Ä
g
Q !Q Q 2ECÄ Ä53,"Ä]¢
2K" UÄ ¨~¢Ä !"Q ¬Ä ¢¬¨Ä ¢Ä~Ä ¨Ä¥Ä¨~®¬Ä Q
!Ä t®¦¦¥¬Ä~Ä "Q !QQ ?Ä
Ä ~¢ÄÄ~¨~Ä ¥Ä !"Q ¬ÄµÄ9!Ä ]¢Äį~®Ä¥Äµ7) 2M" lÄ 1Q ~¢Ä Ĩ~Ä¢®¨¬Ä¬®Ä~ 1Q K
Ä~¢Ä 6ÄQ 1 2E"Ä
]¢Äį~®Ä¥Ä 1Q Ä
52# ]¢Äį~®Ä ¥ÄĦ¥¬¯Ä¢¨Ä 2Ä 2Ä
Ä"""Ä Ä
Ä £ ÃÄ S ÀÄ À ÁÄ Á ÂÄ 2Ä
Q
2Ä
2:Ä
2+"Ä
Q E+¾
55 lÄTÄ~¢ÄVÄ Ä±¥Ä¦¥¬¯Ä¦©Ä¢©¬Ä ¬®Ä~ 2Ä 2Ä TÄ VÄ
2M5Ä 5++B6Ä 5++?6Ä
\¢Äį~®Ä¥ÄV Ä 5:$ e¢Ä TVXÄTVÄNÄ TXÄ ? N Ä:Ä ~¢Ä mÄ ¬Ä Ä ¦¥¢Ä ¥ÄVX&Ä Z¬ ~Ä ¦¥¢Ä ¥¢ÄTV ~¢Ä ^Ä ¬Ä ~Ħ¥¢Ä¥¢Ä TXĬ®Ä~ÄTZÄ NÄ T^Ä 5ÄNÄ2%Ä eĬÄ~¬¥Ä¯¢Ä~ÄZÄ ^Ä ~¢ TmÄ ¢©¬Ä~Ä _ıÄ_^Ä H5Ä
Ä~¢Ä _ZÄ ³Ä
&Ä \¢Ä į~®Ä¥Ä³%
5? eĬį¢Ä ~ijÄ
2
=% \¢Äį~®Ä¥ 5 h :Ä
5B lÄ ; ~¢Ä ÄÄ¢¬Ä ¥ÄÄ©Ä ¬¬Ä ¥Ä~Ä ©~¢&Ä t®¦¦¥¬Ä ~¢Ä ; ~© Ä©¥¥¬Ä¥ÄÄ §®~¥¢
~¢ÄÄ~©¬Ä~¢Ä¥ÄÄ©~¢Ä ¬Ä \¢Äį~®Ä¥Ä³%Ä 5E \¢Ä Ä~©¬Ä©~Ä¢®©Ä³Ä¬®Ä~ ³6Ä Q ³ 2Ä Q ³6 ³ 2{Ä 5Ä
:B³ 5B+ÄÄ
5H `¥±Ä ~¢¸Ä ±~¸¬Ä ~¢Ä Ä ±¥©Ä nUv`[oUveWtÄ Ä ¦~©¥¢Ä ¬¥Ä ~Ä ~Ä ¦~© ¥¢~¢¬Ä ~Ä ~¬Ä ¥¢Ä ¯¥±RÄ \¥©Ä¶~¦Ä nUv`[nUveWtÄ nUv`[nUveWt nUv`[nUveWtÄ~¢ÄnUv`[nUvcWtÄ~©Ä¬®Ä¦~©¥¢¬&
2?Ä
5K-Ä W¥¢¬¨Ä~Ĭ§®¢Ä ¥Ä¨~Ä¢®¨¬Ä »}¡¼Ä ¢Ä¸Ä
2Ä \Ä Ä¯~®Ä¥ 5....../Ä }6¤¤ª
5M-Ä \¥¨Ä ~Ä ¦¥¬¯Ä¢¨Ä Ä ±Ä±¨Ä
±¨Ä},Ä}4Ä -ÄÄ}Ä~¨ÄÄ¥¢¬Ä¥ÄĦ¥¸¢¥~ Ä \¢ÄĬ~¬Ä¦¥¬¬ ¯~®Ä¥ÄÄ},ÄQ }4ÄQ }6ÄQ /Ä/Ä/Ä Q }Ä ¬Ä¯¬Ä¸Ä 5..D0
:.1Ä \¢Ä Ä~¨¬Ä¦¥¬¯Ä¢®¨Ä³Ä¬®Ä~Ä 4 5³; ³6 ³ÄQ2 Ä
2"
:2% `¥±Ä ~¢¸Ä ¥¨¨Ä ¦~¨¬Ä¥Ä¢¨¬Ä ³Ä·Ä ¬~¬¸Ä ħ®~¥¢ ³6ÄQ ·6Ä
5³ÄQ ·ÄQ ³·RÄ
:5-Ä \¢Ä Ä ¢®¨Ä ¥Ä ¥¨¨Ä J®¦¬Ä ¥Ä ¦¥¬¯Ä ¢¨¬Ä }4}6 }; }@ }« }F }I ~Ä~¯Ä¥Ä¥ÄÄ¥¥±¢Ä¦¨¥¦¨¬NÄ ] }¡ÄQ }¡Ä }¡ 6Ä ¥¨Ä2Ä DÄ~¢ ] }FÄ 5..D-Ä
::0Ä lÄ Ä~Ħ¥¬¯Ä¢¨Ä¬®Ä~Ä ¥¢Ä¥ÄĨ¥¥¬Ä¥Äħ®~¨~Ä §®~¥¢Ä ?³6 ?°>ÄQ ?³ÄQ °> 5?Ä
.Ä
¬Ä~¢Ä ¢¨#Ä \¢Äį~®Ä¥Ä :?& W¥¢¬¨ÄĬ®~¢¥®¬Ä §®~¥¢¬
³·Ä ³ºÄ ³º ·ºÄ Q
5DDÄ 55?&Ä
\¢ÄÄ¢®¨Ä¥Ä¥¨¨Ä¨¦¬Ä¥Ä¦¥¬¯Ä¢¨¬Ä ]³Ä·Ä ] ~Ĭ~¬¸ÄÄ~¥¯ ¬¸¬Ä¥Ä§®~¥¢¬"Ä :D0Ä \¢Ä Ä¥~Ä ¢®¨Ä ¥Ä¦¥¬¯Ä ¥®¨Ä ¢¨¬Ä ; ¬~¬¸¢Ä ¥Ä ¥ÄÄ¥½ ¥±¢Ä ¦¨¥¦¨¬OÄ ] ; ¬Ä¯¬Ä ¸Ä JÄ ~¢ ] ±¢Ä Ä ¨¬Ä ~¢Ä ~¬Ä ¬Ä ¥Ä ; ~¨Ä ¢¨~¢Ä Ä ¨¬®¢Ä ¦¥¬¯ ¢¨Ä ¬Ä ~¬¥Ä ¯¬Ä ¸Ä J¿Ä q¥Ä ~Ä Ä ¨¬®¢Ä ¢¨Ä ¢Ä ¢¥Ä Ä ~ ¥®¨Ä¢® ¨" 2DÄ
+>H9.NKQ7] %.T;7E.T>1.B] 'BWEN>.4] +%'] *6G=JP] *60S=JG] *M60=-A] )JVG3 +M:+ 25:+Ä
t~®¨~¸Ä MÄi®¸Ä 5++BÄ 33).3; 1; )+8; /513#,+1; 1; 8,5; +; ,; '5'3,01; 0; '',7; ",7; 3"; 13.1; #+; 8,50; '5'3#,+1;
"; /513#,+; 00#1; Q )0&1;
2( vÄ ¬Ä ¥Ä ~Ä :Ä ¢®¨Ä ~¨Ä¢¨~¢Ä ¬¥Ä ~Ä ¢¥¢Ä ¥Ä Ä ¬Ä ¨~¢ ¨Ä¥¨¢~Ħ¥¬¥¢¬$Ä vĨ¢Ä¥Äı¥Ä¢®¨¬Ä¬Ä~Ä5Ä¢®¨Ä~¢ ¬Ä~Ħ¨Ä¬§®~¨$ 5% W¥¢¬¨Ä Ä ¢¥¢¥¢¯´Ä§®~¨~¨~Ä !"#Q ±Ä +"Q 2K+¾$Ä jÄ Ä ¬Ä #" ´¢ÄÄ !Q ~Ä'Q ~¢ÄĬÄ!"Q ´¢ÄÄ #Q ~Ä % Q UĢĢ¨¬¬ Ä¢¨¥¨¬Ä¥ÄĬ¬Ä ! Q $Q !"Q "#Q ~Ħ¥¢¬Ä *QkÄ) Q aÄ ¨¬¦¯¸#Ä s¨¥¯ ~ÄÄ #(Q "'Q ~¢Ä !)Q "&Q ¢Ä *)Q ak$
:$ juÄÄ~Ĭ®¬Ä¥Ä »2Ä5Ä:Ä% $ ÄÄ5?¼Ä ±ÄubÄ 2+$Ä t¥±Ä~ÄuÄ~¬Ä±¥Ä5¢ ¬®¬¬Ä »Ä 8 Z] ~¢Ä »Ä Z] ¬®Ä~Ä Q NQ Q
?% Y¨¢Ä±¨Ä¨Ä´¬¬Ä~Ä ¦¥¬¯Ä¢¨Ä ¬®Ä~Ä @Q ¢¬Ä±Ä5++B' i®¬¸Ä ¸¥®¨Ä ~¢¬±¨
2EÄ
Icm]Rtpw[ CR|_[hR|cURf EfhtcRX
HZlbovHZT{bolHoe}{boly ' 6nw* 9 IcnRV(""/
NnT(""7
NtVpTTn{mPVtwwpNtV(""/)!!9
NnT(""7)!!1
K`{w(""/)!!9
(""7)!!1
cw V|Vn NnTcycwTc|cwcPiVPy`VwmNiiVwyqtcmVn{mPVt( ( 6nw) 6 (""/)
(
+; (""/
'<
D
:""
:""
D
(""""
, 6nw) : IcnRVy`VVr{Nycpn(})
:}
rD
"
`Nw pnipnVwpi{ycpn cywTcwRtcmcnNny
:)
-
+; (
*;
+rD
" K`{wrD
: - 6nw* 8 FnVVNwciR`VRgwy`NyYptw
( '"
w
}ciiVnT }cy` y}pVtpw 6iwp '
(
,
-
/
6
7
:
<
D
'
(
6
(-
'("
7("
/"-"
-",("
,6(::"
-"<'',
K`{wy`Vw{mpYy`ViNwyy}pTc_cywcw'
,D-
/ 6nw* 8
6 6nw+ 6 IcnRV (
## /""()!!1sxm2##*
)!!1
D
/"ynzm3##+
)!!1* 3
(""/)!!1
( }V`N|V( ( >VnRV (D
(
(D
(D
(
2; '(
D
(-
K`{w (D
':"
(
D
':"
':"
(-
(-
'(
,6
'7
'(
K`Vn
8 6ow0 ; Lwbo_6tby`mVybRDVNo
=VpmVytbRDVNo }V`N|V A AW A8[ % &
& G ( ( A8[ AW &
GW(
A8[ * 2; 2AC
GW A8[ 2 2 AC }`bR` bw wNybw]VT }`Vo G ( NoT2;
K`{wy`VwmNiiVwyqpwwbPiV|Ni{Vbw GW/
;r{Niby bw NR`bV|VT }`Vo
2;
AW
[, 6ow* ; MV wbmqi tVqiNRV P NoTP yp_Vy
C ' A C A ( A A ;2 C
; 6ow) 8 CA
RP \( CPA RP
CE PA C CE E EA E EA wboRV $; 8, (
A8- 6ow) 9 2 R CA E E CC2 C 678; E CA 345;
$A E A E
CE-
K`VtVZptV y`V qpboyw bo y`V 9NtyVwbNo qiNoV wNybwZbo_ y`V Vr{Nybpo Rpowbwy pZqpboywpoy`VRbtRiVRVoytVTNy AA NoTpZtNTb{w C,
}`Vncycwy`Vqpcnycny`VYp{ty`s{NUtNnyken_pny`VicnVqNwwcn_y`tp{_` 88 NnU BA / @npy`Vt }ptUw $%(
3 J 'A J NnU E% ( a B ) w E B w EaWRw,
BB, 6nw) B]G
C kI 8 3 v CkI ExE3 RR kIG E 3 YR kI E 3 R]TW kMR[ B]G3]/
K`{w y`V_tVNyVwycnyV_VtiVwwy`Nn C 08 cwA]GBC, 6nw) W8 8 Hy`N_ptNw K`VptVm JL
a ATWE 3 WT3 BVy y`V qVtqVnUcR{iNt Ytpm L
ypIJ
mVVyIJ
NyN
K`Vn 4IJL
4IL
N
K`{w
LN
JL
( WT ) a ( 45 ) 3 ) ( ( aW8/ BTW BW] IL
IJ
AG/ 6nw) Y BVy{ a G CY NnU } 3 T WCT wpy`Nyy`VVs{NycpnPVRpmVw {E}E a { } E ; C{}a8 ; { 3 8 pt } 3 8/
M`Vn { 3 8 }V`N|V G CY 3 8 ; $( aG- M`Vn }a8 }V`N|V T WCTa8 45 3 R/ >VnRV y`Vw{mcw GR 3 Y/ BR, 6nw) BC BVy PVy`VwmNiiVwycnyV_Vt K`Vn}V`N|V
" ( C[8[ ( K`V^twy cnVs{Nicy cmqicVw y`Ny
)
) W /
BR }`ciV y`V wVRpnU cnVs{Nicy cmqicVw y`Ny
$( () BB, AY $( ( AG y`Vn y`VqtpU{RypYy`Vpy`VtcnyV_Vtwcw CF 1; GN 3 CAW3 K`cwcw cmqpwwcPiV wcnRVy`V wmNiiVwy YNRyptpY CBW _tVNyVty`NnAG cw B[ P{yB[E ( GCR #; CBW,K`VtVYptV }V m{wy`N|V$( a BC, K`Vpy`Vty}pcnyV_VtwNtVVNwciwVVnypPV
BG NnU B[1 BT, 6nw* A BVy !) 3 C88R/ K`Vny`VV~qtVwwcpncwVs{Niyp
A]
AW/ 7nw+ W l * a l
J GJ A G A A a a a > G J G J GJG GG GGJ J
K`VtVYptV rI T W rI *R T rM G R x v vl x y v k kI I n vl C G z vl !A C x vl o A z A A A A A A aW, a rI J J J J J
AY/ 7nw( C8R]
Mcy`p{y ipww pY _VnVtNicy }V mN Nww{mV y`Ny e ^ f, Ay Ypiip}wy`Ny A888 _ e ` A8CR NnT N TctVRy R`VRg w`p}w y`Ny A8CR cw y`V pni n{mPVt PVy}VVn A888 NnT A8CR y`NyTc|cTVw RA 0; TE 0; C ? :/ K`{w eaA8CR NnT faA8CT/ K`VtVYptV efaA8CRA 8CTa C8R]/
A[, 7nw) A]C BVy efa Wm fga{ WmNnT gea QWm 8 y`VRpwcnVt{iV
BVy ) PV y`V RVnytV pY y`V RctRiV K`Vn fpi a Ce aAC81 DptVp|Vt pf pgaR Wm K`VtVYptV P y`VRpwcnVt{iV {Ea RE RE C 0; R 1; RAC8a R[d{ a RmO6
K`VnQX a{E RaR[ AWa GC3 K`{w J A A NtVN hefg a 5 QSwcnW8 a 5 0; GC 0; 6 a [ kI7
K`VtVYptV a[ J NnTEa [ J EaA]C6 A]/ 7nw+ R E a OQ AWa [ & Q Q' AWa}E[}AW ";
Wa 8dE Q R Ea8 "; a8 NnT QaR, E}E[}A C8
IcnRVe
9.
cyYpiip}wy`Ny P9:
.
9.
K`{w P,;9.
"
9.
@ny`cwwpi{ycpn }V`N|V{wVTy`VYNRyy`Nyy`Vw{mpYy}pnpnnV_Nyc|Vn{mPVtwcwVtppni}`Vn Ppy`n{mPVtwNtVVtp ("
6nw, ' FPwVt|Vy`Ny
Ipy`VpniqpwwcPicicycwy`NyPa9%
YptVNR`)9%
(
:
>VnRV P2Px9: (%
6nw) %.,
FPwVt|Vy`NyYptNnqpwcyc|Vn{mPVt|
%
N"
A4# 4
|
.
K`{w}V`N|V %
%
%
///
9
K /// 9%1/ >VnRV$ L9%"
:;9%(
:; j%
9%-.
:; j9%., ((
6nw+ (%%
%
%
I
J
%;(
J I
%;(
9 ."";
+; J I 9%;(
'; I
); (; ; (""0)(""-)
I
J
-"";
;
K`VqtcmV YNRyptcNycpn pY-"";
cw _c|Vn P."";
9%;
0; (%%
IcnRV%;( NnTJ
#; I
cyYpiip}wy`NyI 9%;
NnTJ
9(%%
(5
1; ,
(,
6nw) %":
BVyy`VNtVNpYOITS
PV]&
NnTy`VNtVNpYOIMR
PV])
BVyy`VNtVNpYOItlL PV]
wpy`Nyy`VNtVNpY OIJZ
cwNiwp]
wcnRVZ
cw y`VmcTqpcnypYJL
Ep} ITIS
] NnT )
]
IZ IL
K`VtVYptV
])
]r
IT
IR
Ik
IL
IZIJ IT IS
,
( '( TR
9
K`VtVYptV EpyVy`Ny ) 9 T
S
8(
(
;
IT IR
IZ IJ
(
(%
, ' C .
8(
;
,
(
|
9%":
CQ, 6nw) C ( a
; - CqJ Ct L a b a Cr , Q G CqJ CqK CqK ;
KaVn C a qJ "; qJ bC !; DQ ; b8 NnT D CqJ ; a 8, Ka{w F QDC qI V CsSP DQ;qI 1P D Cs ; 188CqI sa C, CT. 6nw- ]8 IcnRV{} NvVyaVvppywpYyaVVs{NycpnD ~Q Q ~C c8 cyYpiip}wyaNy {}a ~Q NnT NPcQC /
KaVn{D} Db{}D C{}a ~Q D Z~C yvcNn_iVcw vc_ayNn_iVT NnT 1]80
";
{D} Db~D- >VnRVyaV
CW- 6nw+ CT FPwVv|VyaNy {}j {}
1mN~ { }, 6iwp D #; ; wcnRV D ; 1 D #; "
Ka{w }VaN|V
Ka{wyaVVs{NycpnPVRpmVwDc GTCT8 G
CT ;8 a8 KaVvVYpvV yaViNv_Vwy |Ni{VpY cw CT,
;
bCT ;8,
CY- 6nw+ GW KpTVyVvmcnV yaVn{mPVvpYqNvycycpnw }Vf{wyaN|VypTVRcTVyaVn{mPVvpY }Nw pY cnwVvycn_ N qNvycycpn PVy}VVn y}p RpnwVR{yc|V |p}Viw cY }V cnwVvy N qNvycycpn Ny Nii Ka{wyaVypyNin{mPVvpY qNvycycpnw cw Q 0; G 0; GbGW1 CZ/ 6nw+ ]8;;
; ; ; + bC88Qa C88GC88Q { H { 99S {D99P D D ; C88Q *; C88T c ;C ) G,,,C88Q c; a C88]8;;2 {< C
>VnRV C888888c]8;;,
CC
C]/ 6nw. Q8; 8RpmqNtcn_y`VRpV\RcVnywpY cycwVNwyp wVV y`Ny { 1, BVyycn_ }V_Vy { {<
) {
;; ;C;) G
;
;
;
K`VtVYptV { 9{< { ; - EpyVy`Ny C88T T /; Q8; }`VtV Q8; cw NqtcmVn{mPVt Ay cw VNw yp wVVy`Ny N qtcmVn{mPVt) Tc|cTVw pnicY ) ) K`VtVYptV y`VwmNiiVwyqpwwcPiV|Ni{VpY cw Q8;,
G83 6nw. ; IcnRV y`V V~qpnVny pn y`V iVYy wcTV pY y`V _c|Vn Vs{Nycpn cw npnVtp cY $; 8 cy Ypiip}wy`Nyy`VVs{NycpncwqpwwcPiVpni }`Vn
IcnRV $; 8 cy Ypiip}wy`Ny
;/
G;3 6nw/ W MVRNntVNttNn_Vy`VVs{Nycpnyp_Vy DC D C
8/
6w N s{NTtNycR Vs{Nycpn cnyaV|NtcNPiV y`VTcwRtcmcnNny pYy`VNPp|VVs{Nycpn cw_c|VnP :cwRtcmcnNny 1C D Q ; D C
Q;C GD 1;W G C D-
IcnRV cw Nn cnyV_Vt y`V TcwRtcmcnNny m{wy PV N qVtYVRy ws{NtV NnT y`{w npnnV_Nyc|V >VnRV ;W G ( D ) 8 :;
;W C D ') 7 :; ju
) )
Q C =C
,C
G3
IcnRV cw Nn cnyV_Vt cy Ypiip}w y`Ny cyw pni qpwwcPiV |Ni{Vw NtV _c|Vn P 8;CGQ- M`Vn ;G y`V TcwRtcmNny cw ;C }`cR` cw npy N qVtYVRy ws{NtV K`{w 8(
Q3 M`Vn ( 8 }V`N|VD C 18 <= &; 8( M`Vn C }V`N|VDQ( 8 :; 8Q3 M`Vn Q }V`N|VD WZ 8 >? ( (
Q, K`{w y`VtVNtV W wpi{ycpnw nNmVi 88 (
88CQCCQQQ,
! 6nw) ;GG <tpm c cycw VNw ypwVV y`NyyaV}`piVptTVtVT&y{qiV cwTVyVtmcnVTP {> NnT { 3 AycwVNwypR`VRhy`Ny {V G{>X{ , K`{wP cc }V`N|VG{>T{ C88U, D D D Ay Ypiip}w y`Ny {> cw Tc|cwcPiV P T3 MtcyV {> T/ K`Vn }V `N|V ;TT{ D C88T >? G{ ( Q8;, K`V qpwwcPiV |Ni{Vw pY { NtV C/
Z G]Z cY Ppy` D D
CG
NnU { NtVqpwcyc|V EpyVy`Ny VNR` pY y`VNPp|V|Ni{VwpY { UVyVtmcnVw N {ncs{V E E qpwcyc|V cnyV_tNi |Ni{V pY {? wNycwYcn_ y`V Vs{Nycpn G{@ ) T{ a (""0
K`VtVYptV E y`VtVNtV ;GG w{R`wVs{VnRVw GG/ 6nw) ;C BVy.; PVNncnyV_VtwNycwYcn_y`V_c|VnVs{Nycpn wpy`Ny R|E! RnR|kM CRb8 ;
R|E R| CRbn R| /
IcnRV ~cwcttNycpnNi cy Ypiip}w y`Ny R|E R| CRa "
NnU R|a "
I{Pwycy{ycn_ |b cnypy`V^twyVs{Nycpn }V`N|V
R E R CRb" ;
E R]Wa "
;
" ;C ' a "
;
a ;C '
IcnRV cwNqpwcyc|VcnyV_Vt }V`N|V b; C3 GR3 6nw) C I{PytNRycn_y`VwVRpnUVs{NycpnYtpmy`V]twypnV }V`N|V a CTT# CCRa G; ;
bG;3
IcnRVG; cwNqtcmVn{mPVt }V`N|Vc; NnUbG;5 Kp_Vy`Vt}cy`y`V]twy _c|Vn Vs{Nycpn }V`N|V
;G;!
( CTT
;
$;T ;Y( "
;
a ;T pt ;Y4
M`Vn c;T a; }V`N|V a ;W, M`Vn b;Y c; }V`N|V a ;R, K`VtVYptV y`VtVNtVy}pw{R`wpi{ycpnw nNmVl ;T;;W NnU ;Y;;R,
CR
,/ 6nw* (%" BVydfihE
p\\\d
q\\f
p\g h
PVNYp{tUc_cyn{mPVtUc|cwcPiVP8
AnqNtycR{iNt %
=>
d?@
;
MVNtV_c|Vn y`Nyhfgd
F
q\\\h
q\\f
q\g
d
cx NiwpUc|cwcPiV P 8 FPwVt|V y`Ny dfih hfgd
D
;;;d;;;h
;;; d h
IcnRV 8
UpVw npy Uc|cUV;;;
cy Ypiip}wy`Ny8
m{wyUc|cUVd
h
K`{w d
; h
mpU 8 IcnRV8
Uc|cUVw%""%
}V`N|V %
\\\d
%
\\f
%
\g
h
; "
mpU 8
H
d
p\\f
q\g
h
; "
mpU 8
G
q\\f
q\g
; "
mpU
G
%"%"f
g
; "
mpU 8 H
%
\f
g
; "
mpU 8
K`VtVYptV }Vm{wy`N|VdF
h
mpU8
NnU p\f
g
; "
mpU 8 9pn|VtwVi PtV|Vtwcn_y`V NPp|V Nt_{mVnyw pnVVNwciwVVwy`NycY d
; h
mpU8
NnU q\f
g
; "
mpU 8
y`VnPpy`dfih
NnUhfgd
NtV Uc|cwcPiV P8
Ep} y`VtVNtV%.
qNctwpY d
f
wNycwYcn_d
; h
mpU8
}`cR`NtV %
%
(
(
,
,
;
;
%
' (
;
8
"
'%
;
(
IcnRV"
=>
q\f
g
AB
;;
y`VtV NtV%/
qNctw pY f
g
wNycwYcn_ p\f
g
; "
mpU 8 cnRi{Ucn_ "
"
AnYNRy y`cwcw Vr{Niypy`V n{mPVtpYnpnnV_Nyc|VcnyV_VtwiVww y`Ny%""
y`NyNtVUc|cwcPiV P 8
K`VtVYptV y`VtVNtV%-
0; %/
D
(%"
n{mPVtw }cy` y`VtVr{ctVUqtpqVtycVw
(/
G`k\OqnuY BOz^YgOz`SOd Ddgq`OV GBD
HZlbovHZT{bol HsZTbQeFo}lWHoe}{boly IbnRVy`Vw{mwpYy`VTb_bywpYy`Vy}pn{mPVtwNtVy`VwNmV y`ViVN|Vy`VwNmV tVmNbnTVt }`Vny`VNtV Tb|bTVT P;
K`{w y`VbtTb[VtVnRVbwTb|bwbPiV P;
Ip y`VqpwwbPiVNnw}VtwNtV!% pt' ?ybwVNwypwVVy`NyPpy`RNwVwNtVqpwwbPiV(
K`{w
PXWJ
9; PXUZ
PYW
[
9; PYU
N
P[WR
9; PNLQ
PKLS
9; PZUS
XW
XW WU
ZU
f
[W
d UL
f LW
f
WU
UY
[W
f LW
d
VY
d
ZU
d UL
D{iybqibn_y`VYp{tVr{Nybpnw }V_Vy U
Y
W
U
X
W
U
Y
df
X
W
W
U
X
W
U
Y
df
}`bR`bViTwXW
D
UY
I
Z
[
&
! I{qqpwVy`Nyy`bwtVw{iybwnpyyt{V FPwVt|Vy`Ny]
`NwypyNipY D
#$VjVmVnz w{PwVyw CVy ) ~
!"( ! ( )D
#$ PVy`V#$ViVmVny w{PwVywpY ) NnT )F
!"(
$ ) K`Vn P y`V Nww{mqybpn y`V#$ |Ni{Vw ) NtV m{y{Nii TbwybnRy IbnRV ! ( ) =>
#& NnT y`VtV NtV V~NRyi#$ n{mPVtw Ytpm! y`tp{_`#& }V `N|V
%
~
#!( ) D
- /
D
~,
-
-
7
JcnRV y`VuV cw qNcu w{mmcn_ yp,
NnT N qNcu w{mmcn_yp-7
y`Vn{mPVuw (
(,(-
%; ) 8{yy`Vn
(-
D
(
(,
_c|VwucwVyp NRpnyuNTcRycpn - BVy u
D
BVy Zv
D
u
i }`VuV i cw Nn cnyV_Vu w{R` y`Ny ( i ) GPwVu|V y`Ny cY Zv
D
deihj
}`VuV d
e
i
h
NuV y`VTc_cyw pYZv
y`Vn y`V]uwy-
Tc_cywpYZv o
NuV c deih
cn }`cR`RNwVy`V]Yy`Tc_cycw d
j
#; -; ccy`V]uwyYp{uTc_cywpYdeih
pu ccc y`V]uwyYp{uTc_cywpYdeih
( K`ViNwyRNwVRNn`NqqVnpni }`Vnd D
;
K`VuVYpuV }VwVVy`NycYdC
;
Nmpn_ y`Vn{mPVuw Z!
Z$
Z$!
y`VuVcw NyiVNwypnV Ypu }`cR`y`V]uwy Yp{uTc_cyw NuV y`V ]uwy Yp{u Tc_dyw pY #%&")
}`VuV #%&") NuV y`V ]uwy Yp{u Tc_cyw pY Z! ?y Ypiip}w y`Ny y`V ]uwy Yp{u Tc_cyw pY y`V n{mPVuw n{mPVuw Z!
Z$
Z$!!!! cnRi{TVwNiipY -
:;;;
>VnRVy`VNnw}Vucw Vw
(7
!4=/%B?C.J %F0.8%F4)%7J 7H8B4%+J !J @-<J -(D2><J >G<*J J "7."%)""
U\[s\{[V %Hs\ )""2 AGM5FQ $$Q % DJ5GI:BAG
.F:I5Q OBJFQ /AGM5FGQ :AQ I95Q /AGM5FQ G955IQ CFBL:454Q /A4Q G9/45Q I95Q /CCFBCF:/I5Q 1J11>5GQ 15>BMQ OBJFQ /AGM5FG Q (BQ GI5CGQ /F5Q A55454Q IBQ ;JGI:7OQ OBJFQ /AGM5FGQ /29Q DJ5GI:BAQ 2/FF:5GQ % @/F< (BQ 2/>2J>/IBFGQ /F5Q />>BM54 Q
% Bfs[ ~d\mV{~ ~dz\\[fcf~{t] 7&$$ % ) >fzZm\{?' Vs[ ?* dV\ zV[ff . Vs[ 5 z\{u\Z~f\m Sd\ ZfzZm\{ fs~\z{\Z~ V~[f{~fsZ~ utfs~{ Q Vs[=<utfs~Nt~{f[\?* mf\{ts~d\mfs\[\~\zofs\[X Q Vs[ =V~V [f{~VsZ\t]2]zto~d\Z\s~\z t] ?( Mtfs~Pf{Zdt{\sts?- {t~dV~ OPf{~Vsc\s~ ~t ?* V~PBfs[~d\m\sc~dt] ~d\{\co\s~ NP . Bfs[ ~d\mVzc\{~ ut{f~f\ fs~\c\z r{Zd~dV~r \s[{ f~d \
VZ~m %""\zt{ 0 Bfs[ ~d\mVzc\{~Vm\t]
2! Bfs[ ~d\mV{~~t[fcf~{ fstz[\z t]
4 Bzto~d\bz{~ )""2sV~zVm soX\z{ % t]~d\oVz\VzXf~zVzfm Zdt{\sUdV~f{~d\ m\V{~ Vm\ t] % ~t \s{z\ ~dV~ ~d\z\ f{ V~ m\V{~ ts\uVfz t] soX\z{ {Zd ~dV~ ts\ t] ~d\of{ [ff{fXm\ X ~d\t~d\z; 5 J\~ _$ #% % %$% % $% % %% !$"% d\z\ Vz\ stss\cV~f\ fs~\c\z{ ]tz "% ) r! G] k% )%Vs[ _)2 56.25 bs[~d\Vm\t] _%" !
6 G] $% f{Vz\VmsoX\z ~dV~{V~f{`i\{
%"" %"" $% %
% $% %
)6
% $%%
77 %""
542
aq[ ~d\Vm\t] J"& J E\w\ #JJ [\qt~\{~d\mVwc\{~eq~\c\w #J & Td\ ]qY~etq & e{ [\aq\[ ]tw Vmm ut{e~e\ eq~\c\w & Vq[ ~Vl\ tq qtqq\cV~e\
eq~\c\wVm\{{Yd~dV~ &
& & & Vq[ &
&& J & & & & && &
&
tw
.... <m{t ]twVmm & & &
A\~\woeq\
& & F~e{lqtq ~dV~ ~d\.{e[\{t] V ~weVqcm\ Vw\Ytq{\Y~e\ut{e~e\eq~\c\w{Vq[ ~d\
mVwc\{~Vqcm\e{~eY\~d\{oVmm\{~Vqcm\Beq[ ~d\u\weo\~\wt]~de{~weVqcm\ & <~weVqcm\ QQ e{eq{YweX\[eqVYewYm\t]wV[e{& e~d #QQ
% & <m~e~[\{
Vq[ Q t] Q eq~\w{\Y~ V~ " Q Beq[ ~d\ {oVmm\{~ ut{{eXm\ Vm\ t] ~d\ m\qc~dt] ~d\{\co\q~ " Q
Q
& J\~ % X\ ~d\ {\~ t] Vmm ut{e~e\ eq~\c\w{ < ]qY~etq & % % {V~e{a\{ & J
Vq[ & & <m{t qt ~t t] ~d\ Vm\{ &&& Vq[ & YteqYe[\ Et oVq ~dw\\[ece~ ut{e~e\ eq~\c\w{ & {V~e{] &
& &
& J & ]tw Vmm & &
%
& J\~ & X\Vw\VmVm\[]qY~etq{t ~dV~ "&
"&$ #& $ $&
]twVmmw\Vm qoX\w{ "& Vq[ $ & Beq[
&
& & J\~#J
& Vq[]tw % )
UdV~ e{ ~d\Vm\t]
&!!& & Beq[ ~d\{oVmm\{~~dw\\[ece~ qoX\w {Yd ~dV~e]~d\ ~dw\\[ece~{Vw\# J &JVq[' J
~d\q
& J\~#J
& Vq[#J
#JJ&JJ'JJ#&JJ&,JJ# 'JJ#&'J
& Vq[]tw 9 . m\~#9JX\~d\mV{~~t[ece~{t]#: 5J#;"J
UdV~ e{~d\w\oVeq[\wt^#1J #IJJ
Q #IJ d\q e~e{ [ee[\[ X 6
& Beq[~d\{oVmm\{~~t[ece~qoX\w (Q {Yd ~dV~~d\ {o t] [ece~{ t] & [ee{eXm\X &
,7
(Q
e{
%6 Beq[~d\m\V{~r{Yd~dV~d\q\\w~d\\m\o\q~{t]~d\{\~%)rVw\Ytmtw\[ w\[twXm\ ~d\w\VmV{\
e{~ qt~q\Y\{{Vwfm[f{~eqY~ t]~d\{Vo\Ytmtw {Yd~dV~ %% Q %"" )"" UdV~ e{~d\oeqeoo %7 J\~ Vq[X\ut{e~e\p~\c\w{{Yd~dV~ %2% J J J )2% Vm\t] : )" Bfq[~d\oV
eoout{e~e\fq~\c\w r{Yd~dV~ r* % %4" %5" %6" %7"
)% Beq[~d\qoX\wt] ut{e~e\eq~\c\w{ {Yd~dV~ r% )r* % .r/
% )""2r*##3
f{[ee{eXm\X r % )) Df\q~\qL{Vq[~\q%{ dtoVq "%XeqVw{\v\qY\{YVqX\]two\[{Yd ~dV~ qt~dw\\twotw\~dVq~dw\\L{Vw\~tc\~d\w: Btw\
Voum\ "%""%""%"%""%%%"%"%% f{ {Yd V {\v\qY\ X~ ~d\ {\v\qY\ "%""%"""%"%""%%%"%%% [t\{ qt~ {V~f{] ~df{ Ytq[e~ftq ). Fq ~weVqcm\ Q Q )6 QQ )%Vq[ QQ %0 Mtfq~{ Q Vq[ Q Vw\ tq f~d Q 5Vq[ %QQ & Q Bfq[ ~d\m\qc~d t]
)0 Btwutfq~{eq~d\tw[\wQQ Q Q me\tqVYfwYm\f~d~d\\
~\q{ftqt] Q o\\~fqc ~d\ \
~\q{etq t] Q V~ Q Vq[ ~d\ \
~\q{etq t] Q o\\~fqc ~d\\
~\q{ftq t] Q V~ Q K\~ *Q Vq[ ,Q X\ ~Vqc\q~{ ~t~df{YfwYm\f~d uteq~{ t] ~Vqc\qY *Q Vq[ , w\{u\Y~f\mQuut{\ *Q Q 4"Vq[ ,QQ 4.@\~\woeq\~d\m\qc~dt] )2 < u\q~Vctq Q Q e{ fq{YweX\[ fqVYewYm\ t]wV[e{ %"{Yd~dV~ Q f{ uVwVmm\m ~t Q Vq[ Q fq~\w{\Y~{ QQ V~ % Rd\ ~Vqc\q~{ ~t ~df{ YfwYm\ V~ Q Vq[ o\\~ ~d\ \
~\q{etq t] Q V~ V Ytootq uteq~ *Q Quut{\ % *QQ )0 Vq[ &*Q ." Bfq[ %
."
!4=/%B?C.J %F0.8%F4)%7J 7H8B4%+J !J A-<J -(E2><J A-(3$6J >G<*J QW~x[W +Hm +""2
"7"" %.."
HH5?CHQ /GQ ?/AOQ DJ5GH:BAGQ /GQ OBJQ 2/A Q 'BQ 2/=2J=/HBEGQ /E5Q /==BM54Q -9BNQ H95Q GH5CGQ :AQ OBJEQ 2/=2J=/H:BAG Q 029Q DPK5GH:BAQ 2/EE:5GQ % ?/E
% <s gs~\c\x g{{vWx\]x\\g] g~g{st~[gg{gXm\ X /Q ]txWs gs~\c\x /Q %J\~ % X\ ~d\{\~t] ut{g~g\ {vWx\]x\\ gs~\c\x{A\~\xogs\ g~dj{~gaYW~gts ~d\Wm\t]
d\x\ nI [\st~\~d\cx\W~\{~gs~\c\xm\{{~dWstx\vWm~t Btx\
Woum\ n+2I +
+ J\~ !Q X\ ~d\ Y\s~xtg[ t] Q Rdxtcd !Q [xW W mgs\ uWxWmm\m ~t Q Ws[ gs~\x{\Y~gsc ~d\ {g[\{ Q Ws[ Q W~ )Q Ws[ +Q x\{u\Y~g\m J\~ +Q gs~\x{\Y~ !Q W~ Q Ws[ )Q gs~\x{\Y~ !Q W~ Q Mxt\ ~dW~ ~xgWscm\{ Q Ws[ Q Wx\ {gogmWx
J\~ 0Q1Q2Q X\x\WmsoX\x{{W~g{]gsc 0Q1QQ3Q /Q 1QQ 2Q Mxt\ ~dW~ "0Q%1Q Q Q.2 Q QQ0
4Ws[ 01Q16 Q 03Q
7
1 MmWY\ +""2 uths~{ ts ~d\ YgxYo]\x\sY\ t] W YgxYm\ St utgs~{ )Q+Q Wx\ {Wg[ ~t ]txo WuWgx t] A5:81BJEGQ g] ~d\Ydtx[ )+Q {X~\s[{ WsWscm\ t]W~ ot{~ %" W~ ~d\ Y\s~x\ Cgs[ ~d\ |oWmm\{~ soX\x t] uWgx} t] s\gcdXty{
.%
!#"$* &&* "&'* `zudp£ Vdvpdzhd£ [zdl£ `V[#£ Zo£_ogy)£ ]k£ G _y"£ %`
C©6Ä
###`
x£Ä ©vÄ ®vv£Ä R` f
²© Ä ®
vÄ j©®Ä ®
£vvÄ t®©Ä j£vÄ ###`
+`
C©7Ä
0
AG v©Ä Ä ®
vÄ £jtsjÄ j»©Ä xÄ KS` jtÄ K n®
Ä s£sv©Ä [¹Ä
,` jtÄ
vsvÄ
j©Ä v¢²jÄ ¡¹v£Ä ¹®
Ä £v©¡vs®Ä ®
EF,` ?``¹v£Ä xÄ E` ¹®
Ä £v©¡vs®Ä ®Ä O? `¹v£Ä xÄ E` ¹®
Ä £v©¡vs®Ä ®Ä O= ?2, 1,` %4`
1
C©8 Sx
0#8`
W` vt©Ä ¹®
Ä v»js®¾Ä %##` ¿v£© Ä ®
vÄ Ä ®
vÄ ¡£vÄ xjs®£¿j®Ä xÄ W` ®
vÄ ¡£v
xjs®£Ä 2` ss²£©Äv»js®¾Ä %##` ®v©Ä ¹vÄvvtÄ ®Ä ¹££¾Ä jn²®Ä®
vÄ¡£vÄxjs®£Ä C ©svÄ ®Ä ¹Äss²£Ä£vÄ®
j
%##`®v©
Ä f
vIJnv£ÄxÄ ®v©Ä®
j®Ä 2` ss²£©ÄÄ ©Ä ¶v
n¾Ä
C©Ä ®
vÄ xjs®£Ä 2 ss²£©Ä +0` ®v©Ä Ä %##` f
²©Ä ®
vÄ j©¹v£Ä ©Ä jn²®Ä 0##` [¹
T0##N` / qT0##N` T0##N` 0 78 Ä A Q7G %4`G 1Ä A 88 2,` 2` 2.` S®Ä x¹©Ä ®
j®Ä 0##` vt©Ä ¹®
Ä v»js®¾Ä 88`¿v£©Ä f
²©Ä ®
vÄj©¹v£Ä ©Ä 0#8`
0`
C©9Ä
0
N£Ä ®
vÄ v¢²j®Ä ®Ä
j¶vÄ jÄ ©²® Ä
R
²©®Ä nvÄ vj®¶vÄ x£Ä xÄ R
Z,` R`+` Z,` %`
# ` ®
v
Z,` R` A Z`
[,` %` @` Z` Z,`. `q G¾Ä ©¢²j£Ä ¹vÄ n®j R` e¢²j£Ä jjÄ jtÄ ©¶Ä x£Ä Z ` ¹vÄ o®jÄ
bv¹£®vÄ ®
vÄ ¶vÄ v¢²j®Ä j©Ä
+Z,`Rq 0`M£
A
,` A
,A
QC) R`
FC
+` jtÄ ®
vÄ ¾Ä ¡©©nvÄ ©²®Ä ©Ä
f
²©Ä #` :` R` A
B1|£
e²n©®®²®vÄ ®
©Ä
¶j²vÄ xÄ £ ®Ä ®
vÄ ¶vÄ v¢²j®Ä jtÄ ²®¡¾Ä ®
£²
²®Ä n¾Ä n®jÄ
0- 7+ R
0- 7R+ R` G
R` ¹v
0 <
Pvsv Ä U/R 0&`+ q R` 0 R` jtÄ ©Ä U/R 0'` 7q 1` 0` ¹
s
Ä
t©Ä xÄ jtÄ ¾Ä xÄ /R` 0`:`#<` "vÄ /R` 6q -q f
v£vx£v Ä ®
vÄ j»²Ä ¶j²vÄ yÄ /R`©Ä 0 `
2`
C©6Ä
0/`
B
[®vÄ ®
j®Ä 6
%` tÄ %##` [¹ /-$$3 `%-$$3
%` /` tÄ -
$$3` 0R` G /` x£Ä ©vÄ ¡©®¶vÄ ®vv£Ä R `f
²©Ä
PvsvÄ / -
0
4`
C©:Ä
A
6B|£A A
6 A
0/` tÄ %## `
%##0`
fjvÄ j¾Ä ©v®Ä
%##0` ²nv£©Ä
h£®vÄ vjs
Ä ²nv£Ä Ä ®
vÄ x£Ä +H=JL` ¹
v£vÄ IL` #`
jtÄJM`©ÄttÄ f
²©Ä n®jÄ %##0` ttÄ ²nv£©Ä J(`
``J($$1` esvÄ ®
v£vÄ j£vÄ %##/`
ttÄ ²nv£©Ä Ä ®
vÄ £©®Ä +##2` j®²£jÄ ²nv£© Ä j®Ä vj©®Ä ®¹Ä xÄ ®
vÄ utÄ ²nv£©
JM` JO` f
vÄ +H>JL` A` +HBJP` xÄ +H>JL` A +H>JM` f
vÄ ²nv£© %##/`%##0` ``+##2` j£vÄ %##/` ²nv£©Ä ¹
v£vÄ®
v£vÄ ©Ä Ä ¡j£Ä ©²s
Ä ®
j®Ä vÄ xÄ ®
vÄ ©Ä t¶©nvÄ n¾Ä ®
vÄ ®
v£!Ä eÄ ®
vÄ j©¹v£Ä ©Ä %##0 ` j£vÄ ®
vÄ ©jv Ä ©j¾Ä
6`
C©:Ä
2#86`
hvÄ ²©vÄ ®
vÄ xjs®Ä®
j®Ä v¶v£¾Ä ¡©®¶vÄ ®vv£Ä sjÄ nvÄ v»¡£v©©vtÄ ²¢²v¾Ä Ä j¾Ä
+%` ¹vÄ©vvÄ®
j®Ä®
vÄ sv~sv®©ÄxÄ q£j£vÄvj®¶vÄ ®vv¤© ®
jÄ +2 `f
v£vx£vÄ "+2` 67/26` ¹
vÄ ¹£®®vÄ Ä nj©vÄ +2` j¹©Ä ²©Ä ®Ätv®v£v +` C©Ä !%`
®
vÄ svsv®© Ä esvÄ
67/26` 6q8 +A +2`G 2`G`+2
¹vÄn®jÄ r£
7`
C©;Ä
6`G I£G DA6£ PvsvÄ "%#` 2#86`
72/`
££R£ £Ä £Q££q%q x£Ä %'q 4q R`:` 88` esvÄ ®
v£vÄ j£vÄ 78` ®v£©Ä Ä ®
vÄ vx®
jtÄ ©tvÄ xÄ ®
vÄ v¢²j®Ä jtÄ 78` =A Q A 642` A 78`=A 8` ¹vÄ tvt²sv 8` x£Ä 7` [¹Ä ©²¡¡©vÄ £G 7 x£Ä %%` 6q R;`V` jtÄ ckqq ®
j®Ä
S£
G G %q4q :`88` f
vÄ N£©®Ä n©v£¶vÄ ®
j®Ä
££G
7V %# ` G 888 V` 642`
¹
s
Ä ¶v©Ä
7 5q £ G
V
A
8
04`
f
v£vx£vÄ
jtÄ 8
5q £q
£G A
//`
Q jtÄ £G
%#` £v©¡vs®¶v¾Ä
8 ` ¹
s
Ä ¡¾Ä
fvÄ v¢²j®v©Ä vjtÄ ®Ä
0.-q jq A 2.-q jtÄ 2 .+q jq A 3.+q J©v¢²v®¾ Ä ¹vÄ ©vvÄ ®
j®Ä 2.+q jq A 2.-q c&$$jUq 7q2.+ q
PvsvÄ ¹vÄ ss²tvÄ ®
j®Ä
3
C©8Ä
//2
f
vÄ ¶vÄ ¥vj®Ä ¡v©Ä ®
j®
[dqqfq FG [dqq[ fq `²®®Ä dq 7q vj®¶v Ä
fq 7q &q ¹vÄ n®jÄ [ )q FG )[ &!q [ &q7q$!q [v½® Ä ©svÄ
esvÄ
[ +q7q [ ) q &q q7q [ )qq[ &qqm$q
¥Ä
[)q 7q $q jtÄ [ &q ©Ä À
&oq7q m$q
£Ä
&o
¹
v£vÄ [+q A
$q ®Ä x¹©Ä ®
j®Ä [+q 7q &q [¹ Ä ®Ä sjÄ nvÄ ¡£¶vtÄ t²s®¶v¾Ä ®
j®Ä [+fq FG fq x¥Ä jÄ fq C© Ä xÄ [ +fq A fq x£Ä ©vÄ 1A ®
vÄ [+dq A dq x¥Ä jÄ dq FG f"q eÄ ©svÄ [ + q;A ++++q7q [3333q7q++++q®Ä x¹©Ä ®
j®Ä [+fq7q fq x¥Ä fq ++++q TÄ ¡j£®s²j£Ä [ + q;A )$$.q7q)$$.q J©v¢²v®¾ Ä )$$.q7q^ + q(A )$$.q FG ^ ) q)A )$$.qq^ )$$.q FG +[ )$$.q jtÄ ©Ä [ )$$.q )$$.#+q A
//3q ^Ä ®
vÄ ®
v¥Ä
jt Ä
^ )$$.q FG ^ )$$-qq^ &q7q ^ .q*A //2q7q//2q f
v¥vx£vÄ [ )$$.q7q //2q
&$
C©7Ä
&.
Wv®Ä Mq ®
vÄ n©vs®£Ä xÄ ,$G Wv®Ä Dq f
²©Ä D>#E 7q j &*# jqq&"q C©Ä <>#9q 7 <> 7 jj &# j q & q "q
-A
A <.A
C©Ä 9Dq7
#A
9>qE>q¹vÄ
j¶vÄ jj &# jq& q qq j &*q\ jq& q q7qjq& q q e¶Ä :A 7q .q f
²©Ä ®
vÄ ¡v¥v®v¥Ä zÄ ®
vÄ ®¥jvÄ ©Ä -qq. q /q q7q &.q ®
© Ä ®
vÄ ¾Ä ¡©®¶vÄ ©²®Ä ©Ä
+-q
: :2
C©7Ä
:
DMÄY DHs©4)Ã!Ä XiDQMÄYYDKHÄ Pvsv DQÄY DM(©YDQMÄY DHs©4)Ã(©YDKHÄ Y .cs©4)Ã
Y : ©Ä ®
vÄ £jt²©Ä xÄ ®
vÄ s£svÄ f
v£vx£v Ä DQÄY :5 ¹
v£vÄ cÄ %)q
C©7Ä
)).q
[ /qfq [ [ /qfqY [ ) q$£ fq x£Ä jÄ fq ®
vÄ x²s®Ä [q ©Ä ¡v£ts ¹®
Ä ¡v£tÄ -q ©®j£®Ä x£Ä +q ¹j£t©Ä [¹Ä [ / q[ .qY [ 3 q[ -qY %)q jt [+qY %.q j£vÄ x²£Ä t©®s®Ä ¶j²v©Ä f
²©Ä Ä v¶v£¾Ä £²¡Ä xÄ -qs©vs²®¶vÄ ¡©®¶v ®vv£©Ä +q ®
v£vÄ ©Ä v»js®¾Ä vÄ ®
j®Ä ©Ä j¡¡vtÄ n¾Ä {Ä ®Ä )$$."q esvÄ ®
vÄ svs®Ä xÄ ®
£vvt®Ä ¡©®¶vÄ ®vv£©Ä sjÄ nvÄ t¶tvtÄ ®Ä v»js®¾Ä )).q £²¡©Ä xÄ -q s©vs²®¶vÄ ®vv£©Ä vjs
Ä ®
v£vÄ j£vÄ )).q ®
£vvt®Ä ¡©®¶vÄ ®vv£©Ä fq ®
j®Ä ©j®©v©Ä [fqY )$$. q esvÄ
%+q
C©<Ä
.q
ev®®Ä jqY EGY l q ¹vÄ ©vvÄ ®
j®
[ jqjqY [ jqj )[ j qj )jq x£Ä jÄ j"q PvsvÄ [ jqj qY jq x£Ä jÄ jq ev®®Ä EGY jq ¶v©
[ jqjqY [ jql )[ jql )lq x£Ä jÄ jq jtÄ ?A PvsvÄ [qqj qlqY[qqj qjq )lqY jq )lq x£Ä jÄ jq jtÄ lq )$$.q%$$$qY ."q
%-q
C©7Ä
f
v£vx£v
)q
? v
Y v| *v|Y fBv 1A : Bv
|Z |Z
vB88LY
%.q
)$$- )$$+q )$$-q vB88IY5 0 3Y )$$/q)$$. )$$/q
)q :£ v @Y )$$.q +q
:RR N£Ä :77v:7xyYvxyvxxyvyvxy ¹vÄ v®Ä RRxyxyvYxyR2 [®vÄ ®
j®Ä VPeÄVW $q ©svÄ x$£y xzY:W %q y : 0G RR£ jtÄ dPeÄ 1G $"q f
²© ¹vÄ ²©®Ä
j¶vÄ xYyYR0 esvÄ vVW : 1AY :RR©Ä ®
vÄ j©¹v£# C©<Ä
+.q
>O2
C©<Ä
>
[®wÄ ®
k®Ä ®
wÄ ¶k²wÄ xÄ D
tÄ F£ ©Ä q¡w®w¾Ä tw®w¦wtÄ o¾Ä ®
wÄ ¶k²wÄ xÄ
p
A p
$£p
D tÄ >77L£ p
!$£p
"D tÄ H2 tÄ H
w w¢²k©Ä >CG>7>>CG>7>>C2 / 2 ¹
q
Ä
k©Ä kÄ q¾qwÄ xÄ w®
Ä O2 C©Ä ®
wÄ ©²Ä xÄ ®
wÄ ©¢²k¦w©Ä Ä wÄ q¾qwÄ ©Ä 7 tÄ F3£ [¹Ä C77K O 9A GGH>2 f
²©Ä ®
wÄ ¦w¢²¦wtÄ k©¹w¦Ä ©Ä >/ tÄ H2 N²¤®
w¦¦wÄ
f
²©Ä ®
wÄ ©w¢²wqwÄ cE<£
>P2
C©5Ä
C7
Yc%£e4£ h
wÄ e£N£7 gqn+ \Ä ©Ä kÄ \t®Ä ²ow¦Ä ¹
©wÄ ¦©®Ä \ C t®©Ä k¦wÄ H£ ktÄ ®
wÄ k©®Ä ®¹Ä t®©Ä k¦wÄ H, c£ktÄ >7$ e4£ f
²©Ä ®
wÄ ©²Ä xÄ t®© xÄ gqn% \Ä w¢²k©Ä H W, C$£>R) c $£e£A FHc£$£Fe£$£ G f
²©Ä ¹wÄ wwt
Vw®Ä \Ä
cL£g tÄ CT -c $£Gx A &> tÄ KU HvFe£A)> tÄ >P2 c£A >
k©Ä Ä ©²®Ä o²®Ä c£A Ge£ H£©ÄkÄ ©²®#
h
wÄ e£
c
>Q2
7 ®
wÄ ©²Ä xÄ t®©Ä ©Ä H W+ Cq >7$ c£ FHc+ F4£ f
²©Ä c£ ©Ä w¶w$ C ©Ä kÄ ©²®Ä f
²©Ä ®
wÄ k©¹w¦Ä ©Ä C7 ©qwÄ ®Ä ©Ä A GR2
C©5Ä
>>
H(£>7 k¦wÄ ¦wtÄ ktÄ GHKOPF£ k¦wÄ o²w Ä ®
w S®Ä qkÄ wk©¾Ä owÄ q
wqwtÄ ®
k®Ä xÄ > C ®
w¦wÄ tÄ ®Ä w»©®Ä >A 5A xÄ ®
wÄ ©kwÄ q²¦Ä ©²q
Ä®
k®Ä $£>A A 5A e?@Ä
>>2
> > ©²q
Ä ®
k®Ä ®
w¦wÄ tÄ ®Ä w»©®Ä >A 5A f
wÄ >H£ ²©®Ä owÄ xÄ kÄ t}w¦w®Ä q²¦ x¦Ä G2 f
²©Ä ©k¾Ä > ktÄ H£ k¦wÄ ¦wtÄ ktÄ G ©Ä o²wÄ eqwÄ P$£> > H'£ P ²©®Ä ®
wÄ ow o²wÄ [¹Ä CqG$£G A P ©Ä ®
k®Ä C ©Ä ¦wtÄ ktÄ
wqwÄ H A >$£ >$£C ©Ä o²wÄ I²® >>M£G$£H$£H H£$£>$£> kÄ q®¦ktq®# e²¡¡©wÄ ®
w¦wÄ w»©®©Ä kÄ q²¦Ä x¦ÄÄ A
xÄ ®
wÄ ©kwÄ q²¦Ä ©²q
Ä ®
k®Ä $£$£ >A
5A
>R2
C©=Ä
G
f
wÄ w¢²k®¾Ä qkÄ owÄ ®¦k©x¦wtÄ ®Ä
G7C CK 2 C77 'A C77 ·¸wÄ ¦©®Ä wwtÄ ®Ä ®
wÄ ²Ä ¡©®¶wÄ ®ww¦Ä ©²q
Ä ®
k®Ä ®
wÄ ®w¦¶kÄ q®k©Ä kÄ ®ww¦#Ä f
©Ä ²Ä ¶k²wÄ xÄ ©Ä C2 h
wÄ q®k©Ä ¾Ä wÄ ®ww¦Ä ®
k®Ä ©Ä G/ f
²©Ä ®
wÄ k©¹w¦Ä ©Ä G/
C7/
A C ®
©Ä ®w¦¶k
G7HRR Vw®Ä {Ä M£ >O7 7A >P7 8A >Q7 9A >R72 ^o©w¦¶wÄ ®
k®Ä {Ä >PKE >KD>PKD' KD2 Vw® A >PKD#>CK2 f
w {Ä A #qq qq X D(>7777A D2 eqw ¼Ä >PKD>CK C©5Ä
GO
jtÄ {# ) >BYB) >777*
# >\8q C) >777> ºvÄ
j¶vÄ { A ) >B6 f
²© > >PKB) >CO8q G7HRR3
®
vÄ ²Ä ¡©®¶vÄ ®vv¥Ä ©²s
Ä ®
j®Ä B A { ©Ä %
C>2
>O Wv®Ä C77KB88L8q { +> v º
v¥vÄ { ©Ä jÄ ¡¾jÄ xÄ 2 SxÄ Y > C77K8q C77K6A >77G0 ^n©v¥¶vÄ ®
j®Ä C77KB88L © ºvÄ
j¶vÄ vY>C t¶©nvÄ n¾Ä ) > xÄ jtÄ ¾Ä xÄ vYC77K 6A >77G ©Ä t¶©nvÄ n¾Ä , >2 f
vÄ ¡¥v xjs®¥¿j®Ä xÄ v ©Ä vYK 7A H7> 6A >P 7A CH5£ eÄ v
j©Ä ¾Ä CJY >O xjs®¥©Ä f
²©Ä ®
v¥vÄ j¥vÄ >O ¡©©nvÄ ¡©®¶vÄ ®vv¥©Ä 1A ©²s
Ä ®
j®Ä C77KB88L ©Ä t¶©nv n¾Ä # >2 C©5Ä
CC0
C©>Ä
CH7OQ
L©tv¥Ä ®
vÄ©v¢³vsvÄxÄ®vÄ >© Ä f
v¥vÄj¥vÄ vv¶vÄ©¡jsv©Änv®ºvvÄ®ºÄ >©Ä ¥Änvx¥vÄ ®
vÄ vx®©®Ä > ¥Ä jx®v¥Ä ®
vÄ ¥
®©®Ä > 0 N¥Ä vjs
Ä xÄ ®
v©vÄ ©¡jsv©Ä ºvÄ sjÄ ¡²®Ä v®
v¥
®ºÄ _©Ä t²nvÄ _©£ ¥Ä ¾Ä vÄ 7 ©vÄ 7 3 SxÄ ®
v¥vÄ j¥vÄ v»js®¾Ä
} t²nv©Ä _©
®
vÄ ®
v¥vÄ j¥vÄ ¾Ä >7#C} ©vÄ _© Ä f
vÄ ²nv¥Ä xÄ ºj¾©Ä x¥Ä ®
©Ä ®Ä
j¡¡vÄ © Ä ££;;*{£5£ f
IJ©Ä ®Ä
ÄvÄ j©ºv¥Ä -©Ä u ~
~ @ 8 B
>>>>)}YCH7OQ0
L
* } ~[r CG2
C©5Ä
>) C}
>C
Vv®Ä (G 2 G¾Ä s©vÄ ¥²v sªÄGY
C>BCQB# >HB M£ P G£
B#C>H P f³«Ä ;>(q M£ >HBP
jtÄ s©Ä]Y
Y
>HBG CQB,C>B C>H CQ
>> >O
¬Ä ®j®Ä <>8q
-A
A
A
#A
B@ S
%A
@A
!A
CpÄ
=OQL9;>q M£ f´Ä + G
B C>H
P Q
8
+"%GY +!#G'G Ä ¯l®Ä ¯§mµvÄ )*G ±Ä ©svv©%Ä f
v¥vx¥v&
Y
3A
#G
Cs©Ä
C> Q Y -Y>C0 G C P
GP
CH2
C©;Ä
QP
Vv®Ä ®
vÄ s£s²s£svÄ zÄ Nabd vv®Ä ®
vÄ vÄ cd k®Ä e. f
vÄ eabd k£vÄ s Á s¾ss#Ä [¹Ä jc_a k]ba jaed©Ä®
k®Äc_aek£vÄk©Äss¾ ssÄ f
²©
csB c_c]Xcacb cecdktÄ dtBXdbd] dad_ dedc0 f
v£vz£vÄ csBG dtB X cecdd q edc X ceG decd cdB0 L©v ¢²v®¾ Ä cdX G OFBXQP0
&A
2A
$A
CK3
C©7Ä
:F
Wv®Ä *Ä ovÄ®
vÄsv®£vÄ zÄ®
vÄs£sv Ä N£©°Ä ¹vÄ ©
¹Ä ®
k®Ä s`*ÄEUÄ cvÄ ÄkÄs s£sv Ä Lvk£¾Ä
s_*Ä c k£vÄ ss¾ ss#Ä
C©Ä j_sh;
jf_aX j_c1m
F7
¡®©Ä s_G c k£vÄ k©Ä ss¾ssÄ f
²©Ä s_*Ä ]UÄ ckÄ vÄ Ä ®
vÄ s£s²s£sv
zÄ Ns_c0
2A
"A
esvÄ ls_o X R7 so ©Ä kÄ tkv®v£Ä zÄ ®
©Ä s£svÄ ktÄ so X
f
©Ä s£svÄ ©Ä k©Ä ®
vÄ s£s²s£svÄ zÄ Ns_^
FQ
CO # G :7B CO©ÄF7X >F 0
!#"$* &&* "&'* `zudp£ Vdvpdzid£ [zdl£ `V[#£ Xm£^mfw&£ ^mfwb£\j£^w£
%q
hwÄ ©
jÄ ©
¹Ä ®
j®Ä Ä ww£jÄ x£Ä j¾Ä Aq -
* &/G
;
X 4
`q Aq Rq®wÄ ®
j®Ä cqsP£ ]q xÄ jtÄ ¾Ä xÄ ]q ©Ä ®
wÄ j£w©®Ä ®ww£Ä ©²s
®
j®Ä ](`q G 3 SxÄ i_q X m%(` q)(`q q](`nq¹
w£wÄ ]q ©Ä ®
wÄ j£w©®Ä ®ww£Ä ©²s
Ä ®
j® ](`q G ®
wÄ ~£tS£ HR_bpq C©Ä xÄ `q jtÄ /A j£wÄ t©®s®Ä ©¢²j£wÄ x£wwÄ ®ww£© PwswÄ ®
wÄ R_q0A Rheq
N£Ä j¾Ä
eswÄ w¶w£¾Ä ®ww£Ä sjÄ nwÄ ²¢²w¾Ä ¹£®®wÄ j©Ä ](`q ¹
w£wÄ aq ©Ä ©¢²j£wÄ x£ww Ä 2:&/G
m%q.Ä #' Ä qnq
f
©Ä s¡w®w©Ä ®
wÄ ¡£x#
Vw®Ä Sq nwÄ ®
wÄ SG#G:q 9q j£wÄ swj£%Ä Vw¹©wÄ > qd Tq j£wÄ swj£ f
w£wx£wÄ >Cq IHq :9q jtÄ Cq ©Ä ®
wÄ t¡®Ä xÄ Fq JKq <9q jtÄ ©Ä ®
wÄ t¡®Ä xÄ >Tq f
²©Ä CFq HJq STq IHq <:q eswÄ ®
wÄ s££w©¡tÄ ©tw©Ä j¨wÄ
.%Ä N£©®Ä ¹wÄ ®wÄ ®
wÄ sw®£tÄ t¶tw©Ä wjs
Ä wtjÄ Ä ®
wÄ £j®Ä +>.%Ä t¡®Ä xÄ
¡j£jwÄ ®
wÄ ®¹Ä ®£jw©Ä j£wÄ ©j£ Ä
9q
:q
>q
@H£
4G
+
VqWqXq BÄ )Ä ©qvÄ x Ä x£Ä v»j¡v Ä Xq 7q ) Ä ®
vÄ VqqWq8q /q jtÄ VWq 7q 3q ¡¾ ®
j®Ä Vq8q Wq8q +qjÄ q®£jtq®Ä C© Ä VqWqXq ©j®©v©Ä j,q /j(qq3 jq VWXq 7q ) Ä v j +(q 7qVWX\jq eqvÄ j +(q /Ä )Ä x£Ä jÄ jq¹vÄ ©vvÄ ®
j®Ä VqWqrj£vÄ jÄ xÄ ®
vÄ ©jv © Ä jtÄ ®
²©Ä VqWqXq A )Ä j©Ä VqW q qXq q7q/ N£©®¾ Ä
[¹Ä ®
vÄ £j¡
Ä EG 7q j,q
/j(qq3j VWXq
j©Ä ©®j®j£¾Ä ¡®©Ä j®Ä jq8q %q jtÄ jq8q + jtÄ q²®©Ä ®
vÄ ¼j»©Ä j®Ä jq8q VWXq©Ä ®
j®Ä Vq Aq g% qWBq %+q jtÄ Xq Aq +g Nj¾ Ä xÄ Xq A -q®
vÄ VWq8q X +(q A %q©Ä ®
j®Ä VWqWYqVXq A %qq Wq%q #W q F²® jq%q #jq ©Ä q£vj©Ä x£Ä jqAq %qg q f
²©Ä %qq Wqq%#Wq A 3qjÄ q®£jtq® .63@?5G B@=DC8@?G C©Ä Ä ®
vÄ £©®Ä ©²® Ä VqWq jtÄ Xq ©j®©x¾Ä ®
vÄ v¢²j®Ä j,q /j(q 3j VWXq8q ) Ä £Ä jj +(q 7q VWXq e²n©®®²®vÄ jq 7q Xq ®Ä v®Ä XX +(q 7q VWXq £ X +(q 7q VWq [¹ Ä j©©²vÄ Xq G -q f
©Ä ¡v©Ä ®
j®Ä VWq 0Ä + Ä ¹
q
Ä Ä ®²£Ä ¡v© V Wq q A )q ¶jÄ CZOZÄ f
²©Ä Xq A -q jÄ q®£jtq®Ä PvqvÄ Xq A -q hvÄ v® VWX8q XX +(q A -q Xq A )q©qvÄ VqWqqXq7q / N£Ä ®
vÄ v¢²j®Ä Zjq 7q VWXq¹
v£vÄ VWXq A
-qZNq 7q ) Ä Z%q 7q -q ¡v©Ä vÄ £® -qZ+q 7q )Ä ¡v©Ä vÄ £®Ä nv®¹vvÄ %q jtÄ +q Z+q8q ) -q ¡v©Ä vÄ £®Ä nv®¹vvÄ +q jtÄ -
nv®¹vvÄ )Ä jtÄ %q Z%q
Zq-q
H0
Vv®ÄsaÄ nvÄjÄ¡j£ÄxÄv
n²£©Äe²¡¡©vÄ s
j©Ä`qv
n²£©Ä¹
Äj£vÄ®Äv
n²£© xÄ aÄ jtÄ aÄ
j©Ä 9G v
n²£©Ä ¹
Ä j£vÄ ®Ä v
®n²£©Ä xÄ s ¹®
Ä aq G 7G f
vÄ n¾ ¶Ä s ®Ä jÄ ¡®Ä ¶v£¾Ä q©vÄ ®Ä aÄ ®
vÄ ²nv£Ä xÄ v
n²£Ä ¡j£©Ä tv©Ä ® q£vj©vÄ jtÄ ®
vÄ ©v®Ä xÄ v
n²£©Ä xÄ
s jtÄ aÄ qqtvÄ
g
²©Ä n¾Ä £v¡vj®Ä ®
©
¡£qvt²£v Ä ¹vÄ ©vvÄ ®
j®Ä ®
vÄ ²Ä ©Ä j®®jvtÄ ¹
vÄ ®
vÄ ¡®©Ä j£vÄ t¶tvtÄ ® 4A
q²©®v£© Ä
4A
G
+. q
©²q
Ä ®
j®Ä ®¹Ä ¡®©Ä j£vÄ v
n²£©Ä |Ä ®
v¾Ä j£vÄ Ä ®
vÄ ©jv
q²©®v£Ä f
²© Ä xÄ ©Ä ®
vÄ ²nv£Ä xÄ ¡®©Ä Äq²©®v£Ä 8 G ®
v
f
vÄ j©®Ä v¢²j®¾Ä x¹©Ä x£Ä ®
vÄ x¹5Ä RxÄ ®
vÄ j£v©®Ä xÄ ®
vÄÄ© Ä ©j¾Ä A
.2 q
®
vÄ ®
vÄ ©jv©® Ä ©j¾Ä
Wq
©Ä AÄ
Vq©
.1q jtÄ ¡£q(* G aK}£q%* ¹
q
Ä qj
nvÄ ¡£¶vtÄ n¾Ä t£vq®Ä q¡²®j®Ä g
©Ä ²Ä ©Ä j®®jvtÄ ¹
vÄ ®
v£vÄ j£vÄ 13Ä q²©®v£© Ä
).q ¹®
Ä .1q ¡®©Ä jtÄ ,)Ä ¹®
Ä .2q ¡®©Ä
2)Ä