human
Give me a Python program to find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m given the following:
PYTHON3 solution for "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.":
assistant
#Here is Python program to find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m.
 def ans(n,m):
    a=bin(n)[2:]
    b=bin(m+1)[2:]
    x=len(a)
    y=len(b)
    a="0"*(max(x,y)-len(a))+a 
    b="0"*(max(x,y)-len(b))+b
    an=""
    f=0
    for i in range(max(x,y)):
        if(f==1):
            an=an+"0"
        else:
            if(a[i]=="0"):
                if(b[i]=="0"):
                    an=an+"0"
                else:
                    an=an+"1"
            else:
                if(b[i]=="0"):
                    f=1 
                    an=an+"0"
                else:
                    an=an+"0"
    answer=int(an,2)
    return answer
                
for _ in range(int(input())):
    n,m=map(int,input().split())
    print(ans(n,m))