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
import math

if sys.subversion[0] == "PyPy":
    import io, atexit

    sys.stdout = io.BytesIO()
    atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))

    sys.stdin = io.BytesIO(sys.stdin.read())
    input = lambda: sys.stdin.readline().rstrip()

t = int(raw_input())

# python codesforces210710.py < input.txt
for zz in range(t):
    #n = int(raw_input())
    str2 = raw_input()
    n, k = str2.split(' ')
    n = int(n)
    k = int(k)
    #print(bin(n))
    #print(bin(k))
    linen = [int(x) for x in bin(n)[2:]]
    likay = [int(x) for x in bin(k)[2:]]
    ans = 0
    fl = []
    fkt = 1
    extr = 0
    bas = 0
    #print(linen)
    #print(likay)
    if len(linen) <= len(likay):
        if len(linen) < len(likay):
            bas = int(''.join(map(str, likay[0:len(likay)-len(linen)])), 2) * (2 ** len(linen))
            #print(''.join(map(str, likay[len(likay)-1:len(linen)-1:-1])))
            for i in range(len(likay) - len(linen)):
                likay.pop(0)
        for i in range(len(likay)):
            if likay[i] == linen[i]:
                fl.append(0)
            if likay[i] == 1 and linen[i] == 0:
                fl.append(1)
            if likay[i] == 0 and linen[i] == 1:
                fkt = 2 ** (len(likay) - i)
                break
        if len(fl) > 0:
            extr = int(''.join(map(str, fl)), 2) * fkt
        #print(bas)
        #print(fl)
        #print(extr)
        if fkt == 1:
            ans += 1
        ans += bas + extr

    print(ans)
