Correct answer: Rf:f51...Nh3[a]/Nh7[a]/Nf3[a]/Nf7[a]/Ne6[a] 2.Nf3#[A] 1...Ng6/[б]//Ng8/[б]//Nd5[c]/Nc6/[б]//Nc8/[б]/ 2.N:f5#[Б] 1.Bc7? zz 1...N<