Correct answer: Nb41...Be7[a] 2.B:e7#[A] 1...Bc6/[б]/ 2.Qb8#[Б]/Bc7#[C] 1...Bd7[c]/Be8[c] 2.Qb8#[Б] 1...c4[d] 2.Qb4#[D] 1...e4 2.Qh2# 1...Bc4/Bd3/Be2