Correct answer: Qa51...Kf4[a] 2.Qf3#[A] 1...Bh4/Be3/Bf6/[б]//Be7/Bd8 2.Qe3#[Б] 1...c3 2.Q:b4# 1.Qb2? (2.Qe5#) 1...Kf4[a] 2.Qd4#[C] 1...B<