Correct answer: Be51...Ng3[a]/Nf2[a] 2.Nf2#[A] 1...Ne3/[б]//Ne7[c]/Nf4/[б]//Nc3/[б]//Nc7/[б]//Nb4/[б]//Nb6/[б]/ 2.N:f6#[Б] 1...Bg5/Bh4/Bg7/Bh
