Correct answer: Qh51...d5+[a] 2.c:d5#[A] 1...f5+/[б]/ 2.K:d3#[Б] 1...Nc7[c] 2.Bd7#[C] 1...d2 2.Kf4# 1.B:b5? (2.Bd7#[C]) 1...d5+[a] 2.c:d5#[A] 1...f5+/[б]/ 2.K:d3#[Б] but 1...N