Correct answer: Be21.Kd4[A]? zz 1...b6/[б]/ 2.Be2#[Б] 1...Kb6 2.Kc4# but 1...a5[a]! 1.Kc3[C]? zz 1...a5[a] 2.Be2#[Б] 1...Ka5 2.Kc4# but 1...b6/[б]/! 1.Bg
