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.
import traceback
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin = IOWrapper(sys.stdin)
# sys.stdout = IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

#####################################################################

import math
from collections import defaultdict as dfd, Counter, deque
from functools import lru_cache
from itertools import accumulate
from heapq import *
from bisect import *
# from sortedcontainers import SortedList

def geti():
    return int(input())

def gets():
    return input()

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

def getsl():
    return input().split()

def getinps(s):
    inps = s.split()
    m = {'i': geti, 's': gets, 'il': getil, 'sl': getsl}
    if len(inps) == 1: return m[s]()
    return [m[k]() for k in inps]

def get2d(nrows, ncols, n=0):
    return [[n] * ncols for r in range(nrows)]

def get_acc(a):
    return list(accumulate(a))

def get_ncr(n, r, mod=10**9+7):
    num = den = 1
    r = min(r, n-r)
    for i in range(r):
        num = (num * (n - i)) % mod
        den = (den * (i + 1)) % mod
    return num * pow(den, mod - 2, mod) % mod

def get_nc2(n):
    return n * (n-1) // 2
    
def get_npr(n, r):
    if n < r: return 0
    return math.factorial(n) // math.factorial(r) 

def put(a):
    return ' '.join(map(str, a))

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

# while y:
#     x, y = y, x % y

ALPHABETS = 'abcdefghijklmnopqrstuvwxyz'
VOWELS = 'aeiou'
CONSONANTS = 'bcdfghjklmnpqrstvwxyz'

inf = float('inf')
mod = 10 ** 9 + 7

xor = 0
xors = []
for i in range(int(3*10**5)+100):
    xor ^= i
    xors.append(xor)

class Solution():
    def main(self):
        (a, b) = getil()    
        xor = xors[a-1]
        req = b^xor
        return a + (0 if xor==b else 1 if req != a else 2)



try:
    Sol = Solution()
    Test = geti()
    for _ in range(Test):
        ans = Sol.main()        
        print(ans)
except Exception as e:
    print(e)
    traceback.print_exc()