Correct answer: R:d51...Bc3[a]/Rb2[a]/Rb1[a] 2.Re3#[A] 1...Be1/Rc3/[б]/ 2.R:e1#[Б] 1...c4 2.Rd4# 1.Nc8?? (2.Nd6#) 1...c4 2.Rd4# but 1...d4! 1.Bc8