Correct answer: Kc41...Bd3[a] 2.Qh1#[A] 1...e:d5/[б]/ 2.Qe2#[Б] 1.Kc4! (2.Qh1#[A]/Qe2#[Б]) 1...Bd3+[a] 2.Q:d3#[C] 1...e:d5+/[б]/ 2.Q:d5#[D]
