Correct answer: Ng41...Ke4[a] 2.Nc3#[A] 1...Ke6/[б]/ 2.d5#[Б] 1...Be4 2.Qg8#[C] 1...Be6 2.Qg2# 1.Nc3+[A]?? 1...Ke6/[б]/ 2.Qe7# but 1...K:d6! 1.N