human
Can you provide a the shortest possible length of the array Bob wrote given the following:
PYTHON3 solution for "Alice gave Bob two integers a and b (a > 0 and b ≥ 0). Being a curious boy, Bob wrote down an array of non-negative integers with \operatorname{MEX} value of all elements equal to a and \operatorname{XOR} value of all elements equal to b.

What is the shortest possible length of the array Bob wrote?

Recall that the \operatorname{MEX} (Minimum EXcluded) of an array is the minimum non-negative integer that does not belong to the array and the \operatorname{XOR} of an array is the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all the elements of the array.

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 5 ⋅ 10^4) — the number of test cases. The description of the test cases follows.

The only line of each test case contains two integers a and b (1 ≤ a ≤ 3 ⋅ 10^5.
0 ≤ b ≤ 3 ⋅ 10^5) — the \operatorname{MEX} and \operatorname{XOR} of the array, respectively.

Output

For each test case, output one (positive) integer — the length of the shortest array with \operatorname{MEX} a and \operatorname{XOR} b. We can show that such an array always exists.

Example

Input

5
1 1
2 1
2 0
1 10000
2 10000

Output

3
2
3
2
3

Note

In the first test case, one of the shortest arrays with \operatorname{MEX} 1 and \operatorname{XOR} 1 is [0, 2020, 2021].

In the second test case, one of the shortest arrays with \operatorname{MEX} 2 and \operatorname{XOR} 1 is [0, 1].

It can be shown that these arrays are the shortest arrays possible.":
assistant
#Here is the shortest possible length of the array Bob wrote.
 import math
from collections import defaultdict
from heapq import heappush, heappop

DEBUG = True

def log(*args, **kwargs):
    if DEBUG:
        print(*args, **kwargs)
 
def ri():
    return int(input())
 
def rl(f=int):
    return list(map(f, input().split()))
 
def rs():
    return input()

class Solution:

    def __init__(self):
        self.xors = defaultdict(int)
        self.xors[0] = 0
        cur = 0
        for i in range(1, 300001):
            cur ^= i
            self.xors[i] = cur

    
    def run(self):

        a, b = rl()
        
        if self.xors[a - 1] == b:
            print(a)
            return

        diff = self.xors[a - 1] ^ b
        if diff == a:
            print(a + 2)
            return

        print(a + 1)
                
                
if __name__ == '__main__':
    
    t = int(input())
    s = Solution()
    #print(s.run())
    #s.run()
    for i in range(t):
        s.run()
    #    print(s.run())