Correct answer: Ba11...Nb2[a]/Nb6[a]/Nc3[a] 2.N:b2#[A] 1...Nd4/[б]//N:d8[c]/Ne5[d]/N:b4[c]/Nb8[c]/Na5[c]/Na7[c] 2.Ne5#[Б] 1...e2 2.Nd2# 1...e6 2.N
