human
What is a the shortest possible length of the array Bob wrote given the following:
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.
impor
assistant
#Here is the shortest possible length of the array Bob wrote.
t math, os, io, heapq
from sys import stdin, stdout
from collections import Counter, deque, defaultdict
from itertools import product
import bisect

# For fast i/o
# def inp():
#     return io.BytesIO(os.read(0,os.fstat(0).st_size)).readline().strip().decode()
# input = inp

# def println(v, end='\n'):
#     stdout.write(str(v) + end)

def getList(method=int):
    return list(map(method, input().split()))

def getInt():
    return int(input())

def getXor(n):
    r = n % 4
    if r == 0:
        return n
    if r == 1:
        return 1
    if r == 2:
        return n + 1
    return 0

def solve():
    a, b = getList()
    x = getXor(a - 1)

    if x == b:
        return a

    if b ^ x != a:
        return a + 1

    return a + 2

for _ in range(getInt()):
    print(solve())

# x = 0
# for i in range(20):
#     x ^= i
#     print(x)