Correct answer: N:c71.Ne3? (2.Rd4#[A]) 1...Nf3[a]/Ned3[a]/Nc4[a] 2.Nc4#[Б] 1...N:e4/[б]//N:e6 2.Bb4#[C] 1...Nf7/Ng4/Ng6/Ned7 2.Nc4#[Б]/
