Correct answer: Qa71...Kd1[a]/Nh3[c] 2.Qb1#[A] 1...Kf1/[б]//Ne4/Nd1[d]/Nd3[e] 2.Qh1#[Б] 1...Ng4/Nh1 2.Qh1#[Б]/Qb1#[A] 1.K:b4? zz 1...Kd1[a