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 = map(int, raw_input().split())
    if n > m:
        print 0
        return
    b = bin(m)[2:]
    l = len(b)
    a = bin(n)[2:].zfill(l)
    p = b.rfind('0')
    if n == m:
        print 1 << (l - 1 - p)
        return
    ans = 0
    for i in xrange(l):
        if a[i] == '0' != b[i]:
            if p < i:
                print 1 << (l - 1 - p)
                return
            ans |= 1 << (l - 1 - i)
            break
    for j in xrange(i + 1, l):
        if b[j] == '0':
            if a[j] == '1':
                print ans
                return
            if p == j:
                ans |= 1 << (l - 1 - j)
                print ans
                return
        else:
            ans |= (a[j] == '0') << (l - 1 - j)
    print ans

T = int(raw_input())
for _ in xrange(T):
    solve()
