Correct answer: Ba31...d1Q[a]/d1R[a]/d1B 2.Rc3#[A] 1...d1N/[б]/ 2.Re2#[Б] 1...Kd1 2.Bb3# 1.Rbb3? zz 1...d1Q[a]/d1R[a]/d1B 2.Rec3#[C] 1...d1<
