Correct answer: Qg51...Kd4[a] 2.N:e6#[A] 1...Kd6/[б]/ 2.Qa3#[Б] 1...Kb4[c] 2.Na6#[C] 1...Kb6 2.Qe3# 1.Qb2? (2.N:e6#[A]) 1...Kd4[a] 2.Qb6#[D] 1...
