Correct answer: Bb71...Kd3[a] 2.Qe2#[A] 1...c2 2.Q:c2# 1.Bc6? zz 1...Kd3[a] 2.Bb5#[Б] 1...d3[c] 2.Qd5#[C] 1...c2 2.Q:c2# but 1...Kc5/[б]/! 1.Qb2? (2.