Correct answer: Nc21...Bf3[a]/Bg2/[б]//Bh1[a]/Bb7[a]/Ba8[a] 2.Qb5#[A] 1...B:d3[c]/Bc6[d] 2.Rc6#[Б] 1...Nb4[e]/Nc1[f]/Nc3[g] 2.Qc3#[C] 1...Bb6/B