Correct answer: Bc51...Nh4[a]/Nf4[a]/Ne1[a]/Ne3[a] 2.Nf4#[A] 1...Bb6/[б]//Bc7[c]/Bd8[d] 2.Nd:b4#[Б] 1...Nf2[e]/Nf6[e]/Ng3[e]/Ng5[e]/Nd2[e]/Nc
