Correct answer: Nd81...Nb3[a]/Nb7[a]/Nc6[a] 2.R:c4#[A] 1...c6/[б]/ 2.Qd7#[Б] 1...c5[c] 2.Bb2#[C] 1.Bd6? zz 1...Nb3[a]/Nb7[a]/Nc6[a] 2.R:c4#[A] 1...c5[c] 2
