Correct answer: N:b71...Ne2[a]/Nb3[a]/Na2[a] 2.Nb3#[A] 1...e6/[б]//e5/[б]/ 2.Qe4#[Б] 1...d2[c] 2.Qc3#[C] 1...d6[d]/d5[e] 2.Ne6#[D] 1.N:d7? zz 1...Ne2[a]/Nb3[a]/N