Correct answer: Qd51...Bg2/Bf1/Bg4+[a] 2.Qg4#[A] 1...Nf7/[б]//Nc6[c]/Nb7 2.Qc8#[C] 1...f:e5[d] 2.Qg6#[Б] 1...Ne6 2.Qg6#[Б]/Qh7# 1.Kh6? zz 1...<
