Correct answer: Qa51...e5[a]/d5[c] 2.Qg1#[A] 1...Bc5/[б]/ 2.Qf4#[Б] 1...Bb6/Bb8 2.B:b6# 1.Bd8? zz 1...e5[a] 2.Qg1#[A] 1...Bc5/[б]/ 2.Qf4#[Б] 1...Bb6/