Correct answer: Bd61...Kg5 2.Be7# 1.Be1[A]? (2.Ne4#[Б]) 1...Kg3[a] 2.Ng4#[C] 1...B:f4 2.Q:f4# but 1...Kg5! 1.Ng4[C]? zz 1...Kg3[a]/Bg5/[б]/
