Correct answer: Bc51...Bf6[a]/Bd8[a] 2.Nd6#[A] 1...Bg5/[б]//Bd6 2.N:g5#[Б]/Nd6#[A] 1...Bf8[c]/Bc5[d] 2.Ng5#[Б] 1.Bf2? zz 1...Bf6[a]/Bg5/[б]//B