Correct answer: Kc51...Nd2[a] 2.Bc2#[A] 1...Nh4/Nf4/Ne1/Ne3 2.Nf4# 1.Bc3? (2.Ne5#[Б]/Bc2#[A]) 1...Ne1/Ne3 2.Nf4#/Ne5#[Б] but 1...N:
