Correct answer: Qc71...c5 2.Qh1#/Qb7# 1.Qh4? (2.Qd4#[A]/Qc4#[Б]) 1...c5 2.Qh1# but 1...Kc5! 1.Qa7? (2.Qd4#[A]) 1...c5 2.Qa8#/Qb7# but 1...Ke4!
