Correct answer: Na11...d3[a] 2.Ne3#[A] 1...Nb4/[б]//Nb8[c]/Nc7[d] 2.Q:c5#[Б] 1...c2[e] 2.Nd2#[C] 1.Kc2? zz 1...d3+[a] 2.e:d3#[D] 1...Nb8[c]/Nc7[d] 2.Q:c5#[Б] b
