Correct answer: Qf51...b1Q[a]/b1R[a] 2.Qa7#[A] 1...b1N/[б]/ 2.Nc1#[Б] 1...b1B 2.Qa7#[A]/Nc1#[Б] 1.Qf3? zz 1...b1Q[a]/b1R[a] 2.Ra3#[D] 1...b1N