Correct answer: Nd41...Qc1[a]/Qb3 2.Q:c1#[A] 1...Qc3/Qd1/[б]//Qb1[c] 2.B:c3#[Б] 1...Qa2 2.Bc3#[Б]/Qc1#[A] 1.Nc3? (2.Qe1#) 1...Qc1[a] 2.Qf2
