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.
import sys
input = sys.stdin.buffer.readline

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

    a, b = 0, 0
    n2, m2 = n,m
    '''while n2:
        n2 >>= 1
        a += 1'''
    while m2:
        m2 >>= 1
        b += 1

    if n > m:
        print(0)
    else:
        '''k = 0
        first_match = 0
        while 1 << k <= max(n,m):
            if (n>>k&1) and (m>>k&1):
                first_match = k

            k += 1

        mx = n^m
        for i in range(first_match+1):
            mx |= 1<<i

        print(mx)'''

        '''
        first_0_1 = -1
        first_1_0 = -1

        k = b-1
        while k >= 0:
            if n>>k&1^1 and m>>k&1:
                first_0_1 = k
                break

            k -= 1

        while k >= 0:
            if n>>k&1 and m>>k&1^1:
                first_1_0 = k
                break

            k -= 1
        print(b,first_0_1,first_1_0)
        a = ((1<<(first_0_1+1))-1)&m
        a = a&((1<<b)-1) - ((1<<first_1_0+1)-1)

        print(a)
        '''

        final = 0
        for i in range(b-1,-1,-1):
            nb,mb= n>>i&1,m>>i&1

            if nb == 0 and mb == 1:
                final += 1<<i
            elif nb == 1 and mb == 0:
                break
        else:
            final += 1
            # no bit where we can't make anything past, so we can make 1 higher

        print(final)