Correct answer: Qf81...Bf8[a] 2.B:f8#[A]/Q:f8#[Б] 1...Bb4/[б]/ 2.Bc1#[C] 1.Bd2? (2.Qh3#[G]) 1...Bb4/[б]/ 2.Bc1#[C] 1...Bd6[c] 2.Qa8#[D] but 1...B:d
