Correct answer: Be41...Bg4[a] 2.Qh8#[A] 1...Bf3/Bc2/Bb3/Ba4 2.B:f3# 1.Kh3? (2.Q:d1#[Б]/Qh8#[A]) 1...Bg4+[a] 2.Q:g4# 1...Bf6/Bg5/B