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.
from functools import lru_cache
from itertools import count
# cost = lambda i : i and i or 2
def solve():
    t = int(input())
    for _ in range(t):
        n, m = map(int, input().split())
        if n > m:
            print(0)
        elif (m + 1) & m == 0:
            print(m + 1)
        else:
            a = 0
            flag = False
            s = []
            flag2 = False
            l = len(bin(n)) - 2
            for b in range(len(bin(m)) - 2):
                # print(bin(a)[2:])
                if not flag:
                    if m & 1 == 0:
                    # print(bin(a)[2:], b)
                        if (n >> b) & 1 == 0:
                            a += 1 << b
                        flag = True
                else:
                    if b < l and (m & 1) != (n >> b & 1):
                        if not flag2 and (n >> b & 1) == 0: 
                            a += 1 << b
                            s.append(b)
                        elif not flag2 and (m & 1) == 0:
                            # print('??')
                            for i in s:
                                a -= 1 << i
                            flag2 = True
                    if b >= l and m & 1 == 1:
                        a += 1 << b
                m >>= 1
            print(a)
            
            
            
solve()
