Correct answer: Bd51...Bg2/Bf1/Bf5 2.Nf5#[A] 1...b6/b5 2.Nc6#[Б] 1.Bg2? (2.Nb5#) 1...B:g2/Bf5+ 2.Nf5#[A] 1...b5 2.Nc6#[Б] but 1...Bd7! 1.B<