Subarray with Given XOR (IBit)

class Solution:
    def solve(self, A, B):
        rec = {0: [-1]}
        cxor = 0
        count = 0
        for ix, n in enumerate(A):
            cxor = cxor ^ n
            if cxor ^ B in rec:
                count = count + len(rec[cxor ^ B])
            if cxor not in rec:
                rec[cxor] = [ix]
                continue
            rec[cxor].append(ix)
        return count