human
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.":


 ... What is a the shortest possible length of the array Bob wrote?
assistant
#Here is the shortest possible length of the array Bob wrote.
 def xorrange(n):
    if n < 4:
        return [0,0,1,3][n]
    h = 2 ** (n.bit_length()-1)
    r = xorrange(n-h)
    if (n-h)&1:
        r ^= h
    return r

t=int(input())
for _ in range(t):
    m,x=map(int,input().split())
    xr = xorrange(m)
    if xr==x:
        print(m)
    elif xr ^ x != m:
        print(m+1)
    else:
        print(m+2)