YouKn0wWho has two even integers x and y. Help him to find an integer n such that 1 ≤ n ≤ 2 ⋅ 10^{18} and n mod x = y mod n. Here, a mod b denotes the remainder of a after division by b. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.

The first and only line of each test case contains two integers x and y (2 ≤ x, y ≤ 10^9, both are even).

Output

For each test case, print a single integer n (1 ≤ n ≤ 2 ⋅ 10^{18}) that satisfies the condition mentioned in the statement. If there are multiple such integers, output any. It can be shown that such an integer always exists under the given constraints.

Example

Input


4
4 8
4 2
420 420
69420 42068


Output


4
10
420
9969128

Note

In the first test case, 4 mod 4 = 8 mod 4 = 0.

In the second test case, 10 mod 4 = 2 mod 10 = 2.

In the third test case, 420 mod 420 = 420 mod 420 = 0.
1
1000000000 1000000000

1000000000 

import sys
testing = len(sys.argv) == 4 and sys.argv[3] == "myTest"
interactive = False
if testing:
    cmd = sys.stdout
    from time import time
    start_time = int(round(time() * 1000)) 
    readAll = open(sys.argv[1], 'r').read
    sys.stdout = open(sys.argv[2], 'w')
else:
    readAll = sys.stdin.read

# ############ ---- I/O Functions ---- ############

class InputData:
    def __init__(self):
        self.lines = readAll().split('\n')
        self.n = len(self.lines)
        self.ii = -1
    def input(self):
        self.ii += 1
        assert self.ii < self.n
        return self.lines[self.ii]

flush = sys.stdout.flush
if interactive and not testing:
    input = sys.stdin.readline
else:
    inputData = InputData()
    input = inputData.input

def intin():
    return(int(input()))
def intlin():
    return(list(map(int,input().split())))
def chrin():
    return(list(input()))
def strin():
    return input()
def lout(l, sep="\n", toStr=True):
    print(sep.join(map(str, l) if toStr else l))
def dout(*args, **kargs):
    if not testing: return
    if args: print(args[0] if len(args)==1 else args)
    if kargs: print([(k,v) for k,v in kargs.items()])
def ask(q):
    sys.stdout.write(str(q)+'\n')
    flush()
    return intin()
    
# ############ ---- I/O Functions ---- ############

# from math import ceil
# from collections import defaultdict as ddict, Counter
# from heapq import *
# from Queue import Queue

def main():
    x,y = intlin()
    if x > y:
        return x+y
    if x == y:
        return x
    # x < y
    if y%x == 0:
        return x
    if y < 3*x:
        return (x+y)/2
    k = y/x-1
    return (k*x+y)/2


anss = []
for _ in xrange(intin()):
    anss.append(main())
    # anss.append("YES" if main() else "NO")
lout(anss)

if testing:
    sys.stdout = cmd
    print(int(round(time() * 1000))  - start_time)