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 __future__ import division,print_function
from heapq import*
import sys
le = sys.__stdin__.read().split("\n")[::-1]
af=[]
for zorg in range(int(le.pop())):
    n,m= list(map(int,le.pop().split()))
    #answer is x^n where x is smallest number  bigger than m with x&n==n
    #l=[n^k for k in range(m+1,m+n+1)]
    #r=m+1+l.index(min(l))
    if n==0:
        af.append(m+1)
    else:
        mb=0#maximal bit of n privé de m
        for i in range(30):
            if (2**i & n) and not(2**i & (m+1)):
                mb=i
        r=0
        for i in range(30):
            if (2**i & n) or ((2**i & (m+1)) and i>=mb):
                r+=2**i
        #return r,r^n,((m+1)|n)^n,((m+1)|n),r&n
        af.append(r^n)
print("\n".join(map(str,af)))
