You are given two integers n and m. Find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m. Here, ⊕ is the [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

\operatorname{MEX} of the sequence of non-negative integers is the smallest non-negative integer that doesn't appear in this sequence. For example, \operatorname{MEX}(0, 1, 2, 4) = 3, and \operatorname{MEX}(1, 2021) = 0. 

Input

The first line contains a single integer t (1 ≤ t ≤ 30 000) — the number of test cases.

The first and only line of each test case contains two integers n and m (0 ≤ n, m ≤ 10^9).

Output

For each test case, print a single integer — the answer to the problem.

Example

Input


5
3 5
4 6
3 2
69 696
123456 654321


Output


4
3
0
640
530866

Note

In the first test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, 3 ⊕ 3, 3 ⊕ 4, 3 ⊕ 5, or 3, 2, 1, 0, 7, 6. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 4.

In the second test case, the sequence is 4 ⊕ 0, 4 ⊕ 1, 4 ⊕ 2, 4 ⊕ 3, 4 ⊕ 4, 4 ⊕ 5, 4 ⊕ 6, or 4, 5, 6, 7, 0, 1, 2. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 3.

In the third test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, or 3, 2, 1. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 0.
# import collections
# import random
# import math
# from collections import defaultdict
# import itertools
# from sys import stdin, stdout
import sys
# import operator
# from decimal import Decimal

# sys.setrecursionlimit(10**6)

p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def li(): return [int(i) for i in input().split()]
def lli(rows): return [li() for _ in range(rows)]
def si(): return input()
def ii(): return int(input())
def ins(): return input().split()


def solve():
    n, m = LI()
    if m<n:
        return 0
    elif n == m:
        if n%2 == 0:
            return 1
        else:
            c = 0
            while n>0:
                if n%2 ==0:
                    ans = c
                    break
                n//=2
                c+=1

            return 2**c
    else:
        nb = bin(n).lstrip('0b')
        mb = bin(m).lstrip('0b')

        nb ='0'*(len(mb)-len(nb))+nb
        # print(nb)
        # print(mb)
        l = len(nb)
        sb = ''
        for i in range(l):
            if nb[i]<=mb[i]:
                sb+= str(int(nb[i])^int(mb[i]))
            else:
                sb+= '0'*(l-i)
                return int(sb,2)

        else:
            c = 0
            while m > 0:
                if n % 2 == 0 and m%2 == 0:
                    ans = sb[:l-c-1]+ '1' + '0'*(c)
                    return int(ans,2)
                n //= 2
                m //= 2
                c += 1
            else:
                return 2**c


def main():
    for _ in range(II()):
        sys.stdout.write(str(solve()) + "\n")


    #     z += str(ans) + '\n'
    # print(len(ans), ' '.join(map(str, ans)), sep='\n')
    # stdout.write(z)


# for interactive problems
# print("? {} {}".format(l,m), flush=True)
# or print this after each print statement
# sys.stdout.flush()


if __name__ == "__main__":
    main()