human
PYTHON3 solution for "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.":
assistant
#Here is Python program to find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m.
 def solve(n,m):
    if n > m:
        return 0
    nbin = '{0:032b}'.format(n)
    mbin = '{0:032b}'.format(m)

    used= []
    free = []
    for i in reversed(range(len(mbin))):
        if nbin[31-i] == "1":
            used.append(2**i)
        else:
            free.append(2**i)
    # print(sum(used))
    left = m-n+1
    tmp = 0
    tot_sum = sum(free)
    for i in range(len(free)):
        if tot_sum - free[i] + tmp < left:
            # print(free[i])
            tmp += free[i]
        tot_sum -= free[i]
    return tmp

import os
import io
# import time
# a=time.time()
if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    T = int(input().decode().strip())
    for t in range(T):
        n,k = [int(x) for x in input().decode().strip().split(" ")]
        res = solve(n,k)
        print(res)