Correct answer: Na41...Bb6 2.N:e7#[A] 1...e6 2.Rd6# 1.Be4[Б]? zz 1...Bd3[a]/Be2[a]/Bf1[a] 2.N:d3#[C] 1...B:d5/[б]/ 2.Na4#[D] 1...Bb3[c]/Ba2[d] 2
