In Mo's algorithm, we either increase one of the ends, or decrease one of the ends. Both ways, we are interested in counting how many intervals there are in the L-R range with xor sum equal to k. If we take prefix xor sums, we can write this as pref[R+1] ^ pref[i] == k ==> pref[i] == k ^ pref[R+1]. Therefore, we need to maintain counts of all possible values of pref[i] in the interval, so that we can update the answer. Since each time L or R changes, one value of pref[i] gets inserted or removed, we can do this in O(1).