Correct answer: Be81...c:d3[a] 2.Bb3#[A] 1.Rb7[Б]? zz 1...c:d3[a] 2.Bb3#[A] 1...Nc7/[б]//Nd4[c]/Nd6[d] 2.Rb4#[C] 1...N:a3[e] 2.Be8#[D] but 1...Na7! 1.