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Ї(gu)Щƒِ(chng)ҫ۵ǡy(tng)(yu)(sh)(xing)Ŀƺܷ(wn)(gu)ŔࡱýhW\(yn)Ї(gu)(du)Wҫِ(j)

2024-08-05 20:03:03 (li)ԴЇ(gu)?qng)?bo)W(wng)   ݋OԷ   

Ї(gu)?qng)?bo)W(wng)85 2024W\(yn)(hu)ِ^(gu)r(sh)g516r(sh)10֣Ї(gu)(du)2015y12~47ö(jing)Ƶă(yu)F(xin)λӪ(jing)ưһ(gu)ýw(du)Ї(gu)(du)ڱÊW\(yn)(hu)W(wng)Ӿ(xing)ĿϵͻԼڂy(tng)(yu)(sh)(xing)Ŀһķ(wn)F(xin)oஔ(dng)P(gun)ע

ڰW\(yn)(hu)100ӾQِM20qСչ(l)4640Լֵo(j)䛊Z(gu)~s](bo)(New York Post)Ԯ˫@yƵİĴP·Ī˹(Kyle Chalmers)Ԓ(xi)“ڊW\(yn)(hu)γ@ӵijɿ(j)yţ(du)չ(l)¼o(j)䛛](mi)|(zh)ұMȥAñِÿ(g)˶һҌ(sh)@(xing)\(yn)(dng)wԡѽ(jng)߱Mȫ(yng)õġ”

Ӣ(gu)·͸(bo)؈D

(dng)?sh)ؕr(sh)g84Ӿ4×100׻ӾЇ(gu)(du)ԓ(xing)ĿĊW\(yn)“”(gu)(du)Լ|(gu)(du)32746ZӢ(gu)·͸(Reuters)4յĈ(bo)Ќ(xi)F(xin)(chng)g•@Ї(gu)(du)ĊZڽY(ji)(gu)(du)64(li)IJo(j)䛡ِ(gu)xR·ī(Ryan Murphy)f(shu)“Ї(gu)(du)ıF(xin)҂ҪһЩЇ(gu)(du)¾ıF(xin)dzɫ”һ(gu)xֺ·ķ˹(Hunter Armstrong)tP(gun)ע(dng)Ї(gu)xչ(l)գʾ“@(du)(li)f(shu)(hu)һݺܿնY!”

(j)(gu)~sr(sh)(bo)(NYTimes)(dng)?sh)ؕr(sh)g4ڌ(du)Ї(gu)W(wng)xJĵČ}(bo)Ќ(xi)(du)Ї(gu)W(wng)(li)f(shu)@ֻһ(g)r(sh)g(wn)}dž?(dng)賿21qJĊZðW\(yn)(hu)ŮӾW(wng)δ܊Ї(gu)W(wng)ZöW\(yn)Ůν

(gu)~sr(sh)(bo)(bo)؈D

Œ(xi)ڬF(xin)(chng)^W\(yn)(hu)(li)Ї(gu)˺;ס@A@(chng)˜(zhn)䣬S̎Ҋ(jin)eǼtЇ(gu)ԱM׌JĸX(ju)(gu)ِһ߀ԮS¼sƾSQِǰ(du)Jĵu(png)r(ji)“@(xing)\(yn)(dng)ҫ۵֮һ”

DƬ(li)ԴЇ(gu)?qng)?bo) ӛ κ껔z

ĩ(gu)W(wng)Ե“@ƪŒ(xi)úܺãЇ(gu)ŮӾW(wng)?yn)鱳ƫ?jin)f(shu)8qxӖ(xn)y͸@Щ(du)һ(g)21qˁ(li)f(shu)Ƕô˲ijɾͰ!”

Ї(gu)ˮ\(yn)(dng)TϫȫtŮp10_(ti)ِЫ@ýɹ(sh)F(xin)W\(yn)ˮŮp10_(ti)Ї(gu)(du)Bڡ(gu)V˾(ABCnews)ِ(bo)QЇ(gu)^m(x)ԓ(xing)ĿϵĽy(tng)ελيZһ@ҲЇ(gu)ˮ(du)W\(yn)(hu)vʷϵ50KЇ(gu)ˮ(du)ڊW\(yn)(hu)ˮِʷo(w)ǰؙM߹܊ֽһ

(gu)„ܿ(bo)؈D

(j)(gu)„ܿ(Newsweek)(dng)?sh)ؕr(sh)g4?qng)?bo)2024W\(yn)(hu)߀ʣһܵĕr(sh)gЇ(gu)ЙC(j)(hu)(chung)쪄(jing)ưvʷW\(yn)(hu)Ї(gu)(du)ˮ(chng)չF(xin)“y(tng)”@10_(ti)yƵӢ(gu)ˮR˹·(Thomas Daley)ِܲLr(sh)NPRʾ“(껺;)ıF(xin)׺o(w)@@Еr(sh)Ӗ(xn)҂(hu)?ni)ˮ?rdquo;

(bo)߀(sh)r(sh)P(gun)עЇ(gu)(du)ِrr(sh)g5賿Ї(gu)(du)4x100ӾِКvʷԵ؊Zһö⣬Ї(gu)ڂy(tng)(yu)(sh)(xing)Ŀƹѽ(jng)ث@3öƺһöy

Ј(chng)Lj(bo)ؔ(ci)(jng)W(wng)аӛDƬ(qun)Ј(chng)Lj(bo)κýwW(wng)վ߂(g)δ(jng)ڙ(qun)D(zhun)d朽D(zhun)ʽ(f)ưl(f)ѽ(jng)ڙ(qun)ýwW(wng)վD(zhun)dʹÕr(sh)ע(li)ԴЈ(chng)Lj(bo)ؔ(ci)(jng)W(wng)аա`߱λ׷؟(z)
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