Correct answer: Qa81...Bd6+[a]/Ba7 2.N:d6#[A] 1...Qf5+/[б]/ 2.Bf6#[Б] 1.Qc4? (2.Bc2#/Qd3#) 1...Bd6+[a] 2.N:d6#[A] 1...Qf5+/[б]/ 2.Bf6#[Б] but 1...<
