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.
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat, permutations
import math
import heapq

MOD = 998244353
def inp(force_list=False):
    re = map(int, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inpf(force_list=False):
    re = map(float, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inst():
    return raw_input().strip()

def gcd(x, y):
   while(y):
       x, y = y, x % y
   return x

def lcm(x, y):
    return x*y/(gcd(x,y))

def qmod(a, b, mod=MOD):
    res = 1
    while b:
        if b&1:
            res = (res*a)%mod
        b >>= 1
        a = (a*a)%mod
        # print b
    return res

def inv(a):
    return qmod(a, MOD-2)

def query(num):
    print num
    stdout.flush()
    return inp()

def my_main():
    kase = inp()
    pans = []
    cur = []
    for _ in range(kase):
        n, m = inp()
        if n>m:
            pans.append('0')
            continue
        n = bin(n)[2:].zfill(33)
        m = bin(m)[2:].zfill(33)
        # print n
        # print m
        ans = []
        ok = 0
        can = -1
        for i in range(33):
            if ok:
                ans.append('0')
                continue
            if n[i]== '0' and m[i]== '0':
                ans.append('0')
                can = i
                continue
            if n[i]== '1' and m[i]== '1':
                ans.append('0')
                continue
            if n[i]== '1' and m[i]== '0':
                ok = 1
                ans.append('0')
                continue
            if n[i]== '0' and m[i]== '1':
                ans.append('1')
                continue
        # print ok
        if ok==0:
            if can != -1:
                ans[can] = '1'
                for j in range(can+1, 33):
                    ans[j] = '0'
        # print ''.join(ans)
        # print bin(530866)[2:].zfill(33)
        pans.append(str(int(''.join(ans), 2)))

    print '\n'.join(pans)


my_main()
