Correct answer: Ba31...Kc3[a] 2.Qd4#[A] 1...Kb5/[б]/ 2.Ba6#[Б] 1.Ba6+[Б]?? 1...Kc3[a] 2.Qd4#[A]/Bb4# but 1...Bb5! 1.Ba3! (2.Qc5#) 1...Kc3[a] 2
