Write a tail-recursive Prolog procedure diff2 (L, R) to compute the following L is a list of ordered pairs of the form (a, b) where a and b are integers R contains the result when we add up all the u's where u = (ba) for all the ordered pairs in the list L (5) For example Given L = [(4,6), (3,4), (5,4)] R = (6-4) (4-3) + (4-5) = 2 + 1 1 = 2 Why is the program in (a) more efficient in terms of space than an alternative program that is not tail-recursive? (3)