Correct answer: N:f51...Ke4[a]/N:a6[c] 2.Bb7#[A] 1...Kc4/Nc6/[б]/ 2.Be6#[Б] 1.Ne6? (2.Qd4#) 1...Ke4[a] 2.Bb7#[A] 1...Nc6/[б]/ 2.Nc7#[C] 1...N