Correct answer: Qd71...Nf4[a] 2.d4#[A] 1...Nd4/[б]/ 2.f4#[Б] 1...Ng1/Nc1/Nc3 2.f4#[Б]/d4#[A] 1...Nh1/Nh5/Nf1/Nf5 2.Qf5# 1.R:e2? (2.d4#[A]/f4#[Б]) 1...c5/B