Correct answer: Bb51...Kf3[a]/Nf1/Ndc4/Nb1/Ndb3 2.N:f4#[A] 1...Kd5/[б]/ 2.Nge3#[Б] 1...Nab3[c]/Nb7[c]/Nac4[c]/Nc6 2.Bc6#[C] 1...Nf3/f3 2.Nc
