Correct answer: Qc51...Bh4/Bf2[a]/Be1/Bf4[a] 2.Qf4#[A] 1...Nb3/Nb7/Nc4/[б]/ 2.Qb7#[Б] 1...f4[c] 2.Qe4#[C] 1.Qb6?? (2.Qe3#) 1...Bf2[a]/Bf4[
