Correct answer: Qb51.Qh1? zz 1...Kf6[d]/Kd4[a] 2.Qh8#[C] 1...Kf4[c]/Kd6/[б]/ 2.Qh2#[Б] but 1...c4! 1.Qd3? zz 1...Kf4[c] 2.Qg3#[A] 1...Kf6[d] 2.