Correct answer: Nb51...Ne6[a]/Nf7/Nc6[a]/Nb7 2.Nc6#[A] 1...Ng2/Ng6/Nf3/[б]//Nf5[c] 2.Nf3#[Б] 1.Bg2?? (2.Qd5#) 1...N:g2/Nf3/[б]/ 2.Nf3
