Correct answer: Qd81...Bc6/Bc8/Ba6 2.B:c6# 1.Qd2? zz 1...d:c5[a] 2.Bc3#[A] 1...d:e5/[б]/ 2.Be3#[Б] 1...Bc6/Bc8/Ba6 2.B:c6# but 1...Ba8! 1.<
