Correct answer: Nd51...Ng7[a]/Nd6[a]/Nc7[a] 2.Nd6#[A] 1...d3/[б]/ 2.c:d3#[Б] 1...f2[c] 2.d3#[Б] 1...f5[d] 2.Ng5#[C] 1...e:f4 2.Qe1# 1.Qg4? zz 1...Ng7[a]/Nd6[a]/Nc
