ChristianGerard (ChristianGerard) on Twitter

Sexual DeEvolution: Porn Killer of Men? | Part 7: The Problem With Porn & (Relationships) http://wp.me/p1p5q3-5N.

twitter.com/ChristianGerard

Sneaky Pins and Invisible Moves

White to play [FEN"2rq1k2/p2bn2r/5p1p/1p5Q/1B6/ P2P2P1/5PBP/1R2R1K1 w - - 0 31"] I chose the strong 1.Qg6!, attacking the rook at h7 and the pawns on the sixth rank.

www.chesscafe.com/text/heisman136.pdf

English Project

Passages O1 - O5 O6 - O0 P1 - P5 Q1 - Q5 R1 - R5 Speaker FA FE FD FI FA FF FD FI FA FG FF FI FB FC FG FJ Sentences F1 - F5 F6 - F0 P6 - P0 Q6 - Q0 R6 - R0 Speaker FA FB FG. FE FI FC FJ FF FD FH FC FG FF FJ FB FC FJ FH FE FD FH FB FE FH 4. Very-Few-Talker Prompt Files The two VFT subjects were members both of the Many-Talker set and the Few-Talker set.

www.phon.ucl.ac.uk/resource/eurom1/EnglishProject.pdf

NNLO predictions for event shapes and jet rates in electron-positron...

The observable has to be infrared safe, in particular this implies that in single and double unresolved limits we must have ′ O4 ( p1 , ..., p4 , q1 , q2 ) → O3 ( p′ 1 , ..., p3 , q1 , q2 ) for single unresolved limits, for double unresolved limits.

pos.sissa.it/archive/conferences/092/051/RADCOR2009_051.pdf

Dierential Cache Trace Attack Against CLEFIA

Q7 = F 0(RK 0, P(0,1,2,3) )3 ⊕ F 0(RK 0, Q0,1,2,3))3 P3 ⊕ P5 ⊕ Q3 ⊕ Q5 = F 0(RK 0, P(0,1,2,3) )1 ⊕ F 0(RK 0, Q0,1,2,3))1 RK 0 can be determined by testing the 232 possibilities against the 8.

eprint.iacr.org/2010/012.pdf

Сборник задач по специальному курсу элементарной математики - Моденов...

www.booksshare.net/index.php?id1=4&category=math&author=modenov-ps&book=1960&page=284

oncology.tomsk.ru/nii/goskontr/files/gk_p1706_23092009_p1.doc

CGH enh 17pq, 20q 7q, 8q, 13q, 20q, 17q 7q21-22, 8q22-24, 11q14-22, 11q24-25, 20q11.2-12, Xq22- 25, Xq25-28. dim 4q, 5q, 6q, 16p, 17p 1p, 5q, 9р,17p, 18q, 19p 18q11.2-12, 18q12-22, 18q22-23.

www.oncology.tomsk.ru/nii/goskontr/files/gk_p1706_23092009_p1.doc

Free Rider 2 Track - treaw

www.freerider2game.com/play-track/87483/

T.M's Gallery | New Scooter Track! [Archive] - The TrackMill.com Forums

Free MTB Code: -1b r -t 1s,-u 1q -h 2i 3 3j u 4k,t 4j 1q 5q,1n 5p 2k 70 41 8k 56 9r,1l 5p 1p 5q,51 9n 5i 9v,5g 9v 6d ac. 7b ao,9l bs ad c7 be cm,be cl ct d2 eh dc ga df,7a aq 76 ba,78 b5 7c b8 7h be 7k bg 7q bl 81 bo 86 br 8e c3 8f c6,8d c1 8h c0 8o.

trackmill.com/forums/archive/index.php/t-10277.html

The stronger version of Theorem 1 mentioned above reads

The same those arguments we infer that G is (3r {qj } whenever w 2 Qj , where 1 6 j 6 r 2. Combining all 4)`–regular, that all vertices from Qj are adjacent Qj , and that all vertices from Pi are adjacent to all vertices from Pi±1 . The absence of (r + 1)–cliques reveals that there can be no further edges. in G . Hence plain that the partition (P1 , . . . , P5 , Q1 , . . . , Qr 2 ) witnesses G ⇡ Hr...

preprint.math.uni-hamburg.de/public/papers/hbm/hbm463.pdf