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 collections import Counter, defaultdict
from sys import stdin,stdout
import io , os , sys
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
input = stdin.readline


def mapinput():
    return  map(int, input().split())

def listinput():
    return  list( map( int, input().split()))

def intinput():
    return int(stdin.readline())

def strinput():
    return input().strip()

def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    prime[0]= False
    prime[1]= False
    ans = []
    for p in range(n + 1):
        if prime[p]:
            ans.append(p)
    return ans

def writ(ss):
    stdout.write(str(ss) + "\n")

xor = []
a = 0
for i in range(300001):
    a ^= i 
    xor.append(a)


for test in range(int(input())):
    def solve():
        a , b = mapinput()
        mm = xor[a-1]

        if mm == b:
            return a 
        aa = mm ^ b
        if aa ==  a:
            return a + 2
        return a+1
    print(solve())

