Correct answer: Ba51...Kf4[a] 2.Nd3#[A] 1...Kd6/[б]/ 2.Bg3#[Б] 1...Kf6 2.Bc3# 1...Kd4 2.Nf3# 1.Bf2? zz 1...Kf4[a] 2.Nd3#[A] 1...Kd6/[б]/ 2.B