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.
# for _ in range(int(input())):
#     n,k = map(int,input().split(" "))
#     arr = list(map(int,input().split(" ")))
#     ml = 0
#     wa = []
#     mx = -99999999999999999999
#     if n>=100:
#         for x in range(n-100,n):
#             for y in range(x+1,n):
#                 b = (x+1)*(y+1) - k*(arr[x]|arr[y])
#                 if b >mx:
#                     mx = b
#     else:
#         for x in range(n):
#             for y in range(x+1,n):
#                 b = b = (x+1)*(y+1) - k*(arr[x]|arr[y])
#                 if b >mx:
#                     mx = b
#     print(mx)

for _ in range(int(input())):
    n,m = map(int,input().split(" "))

    # Here we have to satisfy n ^ k >= (m+1)
    # and value of k must be as minimum as possible

    p = m +1
    p = bin(p)[2:]
    z = bin(n)[2:]

    if len(z)>=len(p):
        p = p.zfill(len(z))
        # print("hel1")
    else:
        z=z.zfill(len(p))
        # print(len(p))
        # print("hel2")
    k = ""

    # print("z =",  z)
    # print("p =",  p)
    for x in range(len(z)):
        if z[x]=='0' and  p[x]=='0':
            k+='0'
        elif z[x]=='0' and  p[x]=='1':
            k+='1'
        elif z[x]=='1' and  p[x]=='0':
            for y in range(x,len(z)):
                k += '0'
            break
        else:
            k+='0'
    print(int(k,2))
        