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.buffer.readline

# let ans be the smallest int not obtained from N ^ [0, M]
# So, N ^ K = ans here K > M as we cover all xor from [0, M]
# Now we find the min ans 
# From max_bit to 0 if N(i) == K(i) then ans(i) = 0
# if N(i), K(i) == (0, 1) then ans(i) = 1; for ans(i) = 0 we need K(i) = 0 but this dec K which we can't do
# if N(i), K(i) == (1, 0) then ans(i) = 0; here we inc K(i) to 1 and since we increased this bit, all bits
# from i-1 to 0 can be changed to match N(i) which will result in ans(i-1..0) = 0
def solve():
    k = M + 1
    ans = 0
    for i in range(30, -1, -1):
        if (N >> i & 1) > (k >> i & 1): # N(i), K(i) == (1, 0)
            break
        ans |= (N & (1 << i)) ^ (k & (1 << i))

    return ans

test_cases = int(input())
for test_case in range(test_cases):
    N, M = map(int, input().split())
    print(solve())