Correct answer: Qc41...d3[a] 2.Qa1#[A] 1...d5/[б]//Bd5[d] 2.Ng6#[Б] 1...B:f5[c] 2.Q:f5#[C] 1...g4 2.Qf4# 1.Qf2? (2.Qg3#[D]) 1...B:f5[c] 2.Q:f5#[C] but 1...d3[a]!
