Correct answer: Bd21...Nd2/Na3[a] 2.N:d2#[Б] 1...Nd6/[б]//Nd8[c]/Na5[d] 2.Q:c5#[C] 1...N:c3 2.Na3#[A] 1.Qc6? zz 1...Nd2/Na3[a] 2.N:d2#[Б] 1.
