Correct answer: Qf41...d6[a]/Bg7 2.c:b3#[A] 1...Bd6/[б]/ 2.Qh1#[Б] 1...e5 2.B:d7# 1.Qd2? (2.Q:d7#) 1...d6[a] 2.Qc3#[D] 1...Bd6/[б]/ 2.Qg2#[C] but 1...d5! 1.B<