Correct answer: Bg31...K:c1[a] 2.Be3#[A] 1...Kc3/[б]/ 2.Be1#[Б] 1.Bh4? zz 1...K:c1[a] 2.Bg5#[C] 1...Kc3/[б]/ 2.Be1#[Б] but 1...Ke3! 1.Bb6? zz 1.