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.
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())):
    a,b = list(map(int,le.pop().split()))
    def f(x):
        if x==-1:
            return 0
        xo=0
        for lo in range(1,20):
            if 2**lo & x:
                if x%2==0:
                    xo^=2**lo
        if x%4==2 or x%4==1:
            xo^=1
        return xo
    xo=b^f(a-1)
    if xo==0:
        af.append(a)
    elif xo==a:
        af.append(a+2)
    else:
        af.append(a+1)
"""def brute(x):
    xo=0
    for k in range(x+1):
        xo^=k
    return xo
for k in range(10):
    print(f(k),brute(k))"""
print("\n".join(map(str,af)))
