For this problem, you are given a carry-in bit `Cin`, as well as the operands `A` and `B`, just as before. However, `A` and `B` are now two bits apiece, with `A0` holding the bit at position 0 and `A1` holding the bit at position 1, and so on. This still produces `R` as a result, though the result is now two bits large, with `R0` holding the result value at position 0, and `R1` holding the result value at position 1. `Cout` holds the carry-out bit, as before.

As before, fill in the table below. Note the ordering of the inputs. The first 3 have already been completed for you.

Cin A1 B1 A0 Be Cout R1 Re 2b) e e 10 1 2c) e e 1 e 1 2d) e 1 1 | 2e) e 1 0 e 2f) e 1 0 1 | 2g) 0 e 1 1 e 2h) e 1 1 1 | 2i)0 1 0 e 2j) 0 1 0 1 | 2k) 0 1 0 1 e 21) 0 1 0 1 1 | 2m) 1 1 0 e 2n) 1 1 0 1 | 20) 111 e 2p) 111 1 | 2q) 1 e e e 2r) 1 e e 1 | 2s) 1 e e 1 e 2t) 1 e e 1 1 | 2u) 1 e 1 0 e 2v) 1 e 1 0 1 | 2w) 1 e 1 1 e 2x) 1 e 1 1 1 | 2y) 1 1 0 e 2z) 1 1 0 1 | 2aa)1 1 0 1 e 2ab)1 1 0 1 1 | 2ac)1 1 1 0 e 2ad)1 11 0 1 | 2ae)1 111 e