Correct answer: R:b41...Nd4[a]/Na3[a] 2.N:c7#[A] 1...Nc8/[б]/ 2.Q:c6#[Б] 1...Bd4[c]/Be5[d]/Bf6[d]/Bg7[d]/Bh8[d]/Bb2[c]/Ba1[c] 2.N:b4#[C] 1...g4 2.N