Correct answer: Re41...Bc2[a]/Ba2 2.Q:c2#[A] 1...Bd2/[б]/ 2.Q:d2#[Б] 1...Bc3/Be1/Ba5 2.Qb5# 1...Bc5/Bd6/Be7/Bf8/Ba3 2.Qb5#/Qd2#[Б]