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.
#coding=utf-8
DEBUG = 0

if DEBUG==1:
    data = open('sample1.in', 'r')

def readline():
    if DEBUG == 1:
        return data.readline().strip('\r\n')
    else:
        try:
            return input().strip('\r\n')
        except EOFError:
            return ''

def readi():
    return [int(t) for t in readline().split()]

t, = readi()
for id in range(t):
    n,m = readi()
    n = format(n,'032b')
    p = format(m+1,'032b')
    k = [0] * 32
    for i in range(31,-1,-1):
        if n[i]==p[i]:
            k[i] = 0
        elif n[i]=='1' and p[i]=='0':
            for j in range(i,32):
                k[j]=0
        else:
            k[i]=1
    kk = ''
    for i in range(32):
        kk += str(k[i])
    #print(kk)
    kk = int(kk,base=2)
    print(kk)
    
if DEBUG==1:
    data.close()    
