human
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.
Solve the task in PYTHON3.
 ... Give me a Python code to find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m?
assistant
#Here is Python code to find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m.

from sys import stdin, gettrace

if gettrace():
    def inputi():
        return input()
else:
    def input():
        return next(stdin)[:-1]

    def inputi():
        return stdin.buffer.readline()

def msb(i):
    i |= i>>1
    i |= i>>2
    i |= i>>4
    i |= i>>8
    i |= i>>16
    i += 1
    return i>>1

def solve(n: int, m: int):
    m1 = m+1
    bl = max(m1.bit_length(), n.bit_length())
    res = 0
    for i in range(bl-1, -1, -1):
        bit = 1 << i
        if (bit & n == 0) and (bit & m1 != 0):
            res += bit
        elif (bit & n != 0) and (bit & m1) == 0:
            break
    return(res)



def main():
    t = int(input())
    for _ in range(t):
        n, m = map(int, input().split())
        print(solve(n, m))

if __name__ == "__main__":
    main()