Correct answer: Nbd71...Nf3/Ng2/Ned3[a]/Nec2 2.B:f3# 1.Ned7? (2.N:c5#[A]) 1...Nbd3[c]/Ned3[a] 2.Nf6#[Б] 1...K:f5 2.e4# but 1...Na6! 1.Na4
