Correct answer: Bf71...Nb4[a] 2.a:b4#[A]/Rc3#[Б] 1...Nc1/Nc3 2.Rc3#[Б] 1.Qg8? (2.Q:d5#) 1...Nb4[a]/Nc3 2.Rc3#[Б] 1...Q:f5[c] 2.Qc4#[C] but 1...Q