Correct answer: B:d51...Ng2[a]/Ng6[a]/Nh3[a]/Nh5[a]/Nfe2/[б]//Ne6[a]/Nd3[c] 2.Ne6#[A] 1...Nd1[d]/Nce2[d]/Ne4[e]/Nb1[d]/Nb5[d]/N:a2[f] 2.Nb5#[Б]
