Correct answer: Qh71...Be5[a] 2.Nd2#[A] 1...B:c3/[б]/ 2.N:c3#[Б] 1...K:d5 2.Q:f3# 1.B:d4? (2.Nc3#[Б]/Nd2#[A]) 1...K:d5 2.Q:f3# 1...N:e3 2.Nd
