Correct answer: Qc51...Bh3[a]/Bf3[a] 2.Qf3#[A] 1...Bh5/[б]//Be2/[б]//Bd1[c]/Bf5+[d] 2.Qf5#[Б] 1...Nc3 2.Rd2# 1.Q:b4?? (2.Q:b1#) 1...Bd1[c]/B