Correct answer: Bg41...c5[a]/Qe2/Qf1[a]/Qb3/Qa2 2.Nc6#[A] 1...Q:d4/[б]/ 2.Bc7#[Б] 1.R:c4? (2.Bc7#[Б]/N:c6#[A]) 1...d:c4 2.Qe4# 1...d4 2.Bc7#[Б]/Q<