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.
for _ in range(int(input())):
    n,m=map(int,input().split())
    if(m<n): print(0)
    else:
        sn=''; sm=''; nn=n; mm=m+1; ans=''; an=0
        while(nn>0):
            sn+=str(nn%2); nn//=2
        while(mm>0):
            sm+=str(mm%2); mm//=2
        for _ in range(len(sm)-len(sn)):
            sn+='0'
        sn=sn[::-1]; sm=sm[::-1]; mm=len(sm)
        if(2**mm-1==m): print(2**mm)
        else:
            sb='1'*mm
            for i in range(mm):
                if(sb[i]==sm[i] and an<=m): ans+=sb[i]
                else: ans+=sn[i]
                an+=int(ans[i])*(2**(mm-1-i))
            ans=ans[::-1]
            print(an^n)
            
                
