Correct answer: Qb51...Kf4[a]/Kd6[d] 2.Qc7#[A] 1...Kf6/[б]//Kd4[c] 2.Qc3#[Б] 1.R:e7+? 1...Kf4[a]/Kd6[d] 2.Qc7#[A] 1...Kd4[c] 2.Qc3#[Б] but 1...K