Correct answer: Qc81...B:b3[a]/Bc6+/[б]/ 2.Qc6#[A] 1...Bd7[c] 2.Nf7# 1...e6 2.Qc5# 1.Q:a4? (2.Qc6#[A]/Qd7#[Б]) 1...Ke6 2.Qa6#/Qc6#[A] 1...e6 2.Q