Correct answer: Qh11...Kf4[a]/Kd6[d] 2.Qg3#[A] 1...Kf6/[б]//Kd4[c] 2.Qc3#[Б] 1.Qb1? zz 1...Kf4[a]/Kd4[c] 2.Q:e4#[C] 1...Kd6[d] 2.Qb8#[D] but 1...
