Correct answer: Nh51...Be1/B:c3[a] 2.Qh3#[A] 1...Nd3/[б]//Nd7[c]/Nb3/Nb7/Na4/Na6 2.N:h6#[Б] 1...Be2[d]/Bd1[d] 2.Qg3#[C] 1.N:e6? (2.Ne5#)
