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.readline

T = int(input())
for _ in range(T):
    n, m = map(int, input().split())
    answer = 0
    if n > m:
        print(0)
    else:
        nbin = bin(n)[2:]
        mbin = bin(m)[2:]
        nlen = len(nbin)
        mlen = len(mbin)
        nbin = '0' * (mlen-nlen) + nbin
        data = ['0'] * mlen
        flg = False
        #print(nbin)
        #print(mbin)
        for i in range(mlen):
            if nbin[i] == '0' and mbin[i] == '1':
                data[i] = '1'
            elif nbin[i] == '1' and mbin[i] == '0':
                data[i] = '0'
                flg = True
                break
        if not flg:
            #print(n ^ (m+1))
            flg2 = -1
            for i in range(mlen-1, -1, -1):
                if nbin[i] == '0' and mbin[i] == '0':
                    data[i] = '1'
                    flg2 = i
                    break
            if flg2 >= 0:
                for i in range(flg2+1, mlen):
                    data[i] = '0'
            else:
                data = ['1'] + (['0'] * mlen)
            answer = int(''.join(data), 2)
            print(answer)
        else:
            #print(data)
            answer = int(''.join(data), 2)
            print(answer)
