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.
def solve(n, m):
    if m < n:
        return 0
    M = bin(m)[2:]
    N = bin(n)[2:]
    N = ('0' * (len(M) - len(N))) + N

    ans = []
    nstart = -1
    lastmzero = -1
    for i, (mc, nc) in enumerate(zip(M, N)):

        if nc == '1' and nstart < 0:
            nstart = i

        if nstart < 0:
            ans.append(mc)

        elif mc == '1' and nc == '1':
            ans.append('0')

        elif mc == '1' and nc == '0':
            ans.append('1')

        elif mc == '0' and nc == '0':
            ans.append('0')

        else:
            return int(''.join(ans) + ('0' * (len(M) - i)), base=2)

        if mc == '0':
            lastmzero = i

    if lastmzero >= 0:
        return int(''.join(ans[:lastmzero]) + '1' + ('0' * (len(M) - 1 - lastmzero)), base=2)

    return m + 1

T = int(input())
for _ in range(T):
    n, m = map(int, input().split())
    print(solve(n, m))
