Correct answer: Bb51...Be4[a]/Bf3[a]/Bg2[a]/Bh1[a]/Bb7/Ba8[a] 2.Ndb3#[A] 1...Be6/[б]//Bf7/[б]//Bg8/[б]//Bc4/[б]//Ba2/[б]/ 2.Nc6#[Б] 1...Bb3/B
