Correct answer: Bf71...Bg4[a]/Bh3[a]/Bd7/Bc8[a] 2.Ngh7#[A] 1...Bg6/[б]//Be4[c]/Bd3[c]/Bc2[c]/Bb1[c] 2.Ne6#[Б] 1...Bh7/Be6 2.Ng:h7#[A]/Ne6#[Б]