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 defaultdict as dd
import math
import sys
input=sys.stdin.readline
def nn():
    return int(input())
 
def li():
    return list(input())
 
def mi():
    return map(int, input().split())
 
def lm():
    return list(map(int, input().split()))
 
 
 
def solve():
    n,m = mi()
    
    if n==0 and m==0:
        print(1)
        return
    if n>m:
        print(0)
        return
    exp = 40
    
    while 2**exp > m:
        exp-=1

    val = 2**(exp+1)-1
    if m==val:
        print(m+1)
        return
    while exp>=0:
        temp = n
        #print(val,n, (temp>>exp)%2)
        
        if (val - 2**exp) > m and ((temp>>exp)%2)==0:
            val -= 2**exp
        exp-=1
        
    #print(val)
    print(n^val)
    

q=nn()
for _ in range(q):
    solve()