Correct answer: Nc51...Rd4[a] 2.Q:d4#[A]/Re6#[Б] 1...Be7+ 2.Q:e7# 1...b4 2.Nc4# 1...Rc5 2.Re6#[Б] 1.Nc7[C]? zz 1...Rd4[a] 2.Q:d4#[A] 1...c5/[б]//R