Correct answer: Nd71...Ke7[a] 2.Nc8#[A] 1...Kc7/[б]/ 2.N:b5#[Б] 1.Kc3? zz 1...Ke7[a] 2.Nc8#[A] 1...Kc7/[б]/ 2.N:b5#[Б] 1...Kc5 2.Qf8#[C] but 1...b4+!