Correct answer: Qf51...Kd4[a] 2.Qd3#[A] 1...N:d1/[б]//Ne2[c]/Nb1/[б]//Na2/[б]//Na4/[б]/ 2.Q:b3#[Б] 1.Qg2? zz 1...Kd4[a] 2.Q:e4#[C] 1...N:d1/[б]//N