Correct answer: Bc41...Ng4+[a]/Ng8[a] 2.N2:g4#[A] 1...Nd3/[б]//Nd7/[б]//Ne4[c]/Nb3/[б]//Nb7/[б]//Na4/[б]//Na6/[б]/ 2.N:d3#[Б] 1...a2 2.B:b2# 1.Bf3? zz