Correct answer: Qb81...c6[a] 2.Nb7#[A] 1...Nb4/[б]/ 2.Qe3#[Б] 1...Nb6[c] 2.Qa3#[C] 1.B:a6? (2.Qb5#) 1...c6[a] 2.Nb7#[A] 1...Nb4/[б]/ 2.Qe3#[Б] but 1...N
