Correct answer: Nb51...Rc6[a] 2.Be4#[A] 1...R:e6/[б]/ 2.Bc4#[Б] 1.Kb4? (2.Be4#[A]/Bc4#[Б]) 1...Rc6[a]/b5 2.Be4#[A] 1...R:e6/[б]/ 2.Bc4#[Б] but 1...B