Correct answer: Qc11...Nc2[a] 2.Qb5#[A] 1...Nc4/[б]/ 2.Qg6#[Б] 1...d4[c] 2.Nc1#[C] 1...Nc3+/Nd2 2.Q:c3# 1.b7? zz 1...Nc2[a] 2.Qa6#[D]/Qb5#[A] 1...N