Correct answer: Bb61.Be6? (2.Bc8#) 1...Rg8[a] 2.Bb4#[A] 1...Rh8/[б]/ 2.Bc3#[Б] but 1...b4! 1.Bb6! (2.Nb8#) 1...b4/Rg8[a] 2.Bc4#[D] 1...Rh8/[