Correct answer: Nb71...Kd5[a] 2.Bf3#[A] 1...Kb5/[б]/ 2.Qa4#[Б] 1...d5[c] 2.Bd7#[C] 1.Bh3? zz 1...Kd5[a] 2.Bg2#[D] 1...Kb5/[б]/ 2.Qa4#[Б] 1...d5[c] 2.B