Correct answer: Qe51...Bh7[a] 2.Qh3#/Qh4# 1...Rh4 2.Q:h4# 1...Rh3 2.Q:h3#/g:h3# 1.Qd6? (2.g4#) 1...Bh7[a] 2.Qh6#[A] 1...Rh2/[б]/ 2.Qd1#[Б] 1..