Correct answer: Qe71...b4[a] 2.Qd3#[A] 1...b:c6/[б]/ 2.Qc8#[Б] 1...a:b6 2.Q:b6# 1.Qd7? (2.Q:b7#[C]) 1...b4[a] 2.Qd3#[A] 1...b:c6/[б]/ 2.Q:a7#[D]/Qc8#[Б] but 1...a:b6! 1
