Correct answer: Rf51...B:d6[a]/Ba3 2.Nc3#[A] 1...Nd3/[б]/ 2.c:d3#[Б] 1...Bf5[c] 2.Ng5#[C] 1...Bc3/Bc5 2.N:c3#[A]/Nc5# 1...B:d2/Ba5 2.Nc5# 1...N