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.
def process(n, m):
    if n > m:
        return [0, 0]
    n1 = bin(n)[2:]
    m1 = bin(m)[2:]
    n1 = n1.zfill(len(m1))
    n1 = n1[::-1]
    m1 = m1[::-1]
  #  print(n1)
   # print(m1)
   # g = m-n
   # if (n^(g+1))==n+g+1:
   #     return g+1
    answer = 0
    is_good = None
    for i in range(len(m1)):
        if n1[i]=='1' and m1[i]=='0':
            is_good = i
    if is_good is None:
        first_zero = None
        for i in range(len(m1)):
            if m1[i]=='0':
                first_zero = i
                break
        if first_zero is None:
            return [2**(len(m1)), 0]
        answer = 2**(first_zero)
        for i in range(first_zero+1, len(m1)):
            if m1[i]=='1' and n1[i]=='0':
                answer+=(2**i)
        return [answer, 0]
    #so this seems to be fine
    
    for i in range(len(m1)):
        if i > is_good and n1[i]=='0' and m1[i]=='1':
            answer+=(2**i)
    return [answer, 1]

t = int(input())
for i in range(t):
    n, m = [int(x) for x in input().split()]
    print(process(n, m)[0])