Correct answer: Qc71.Qe7? (2.Qe3#) 1...Kd2[a] 2.Qe2#[A] 1...Kd4/[б]/ 2.Qe4#[Б] but 1...Nc3! 1.Qb6?? (2.Qe3#) 1...Kd2[a] 2.Qd4# but 1...Nc3!
