Correct answer: Bg71...Bg7[a] 2.B:g7#[A]/Q:g7#[Б] 1...Bh8 2.Q:h8# 1.Qg7[Б]? (2.Q:b2#) 1...B:g7[a] 2.B:g7#[A] 1...Bc3/[б]/ 2.Q:c3#[C] 1...Bd4[c]