human
What is a Python code to find an integer n such that 1 ≤ n ≤ 2 ⋅ 10^{18} and n mod x = y mod n given the following:
Can you solve the below in PYTHON3?
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.
assistant
#Here is Python code to find an integer n such that 1 ≤ n ≤ 2 ⋅ 10^{18} and n mod x = y mod n.

import bisect
import heapq
import math
import collections
import sys
import copy
from functools import reduce
import decimal
from io import BytesIO, IOBase
import os
import itertools
import functools
from types import GeneratorType

#
# sys.setrecursionlimit(10 ** 9)
decimal.getcontext().rounding = decimal.ROUND_HALF_UP

graphDict = collections.defaultdict

queue = collections.deque

################## pypy deep recursion handling ##############
# Author = @pajenegod
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        to = f(*args, **kwargs)
        if stack:
            return to
        else:
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        return to
                    to = stack[-1].send(to)

    return wrappedfunc

################## Graphs ###################
class Graphs:
    def __init__(self):
        self.graph = graphDict(set)

    def add_edge(self, u, v):
        self.graph[u].add(v)
        self.graph[v].add(u)

    def dfs_utility(self, nodes, visited_nodes, colors, parity, level):
        global count
        if nodes == 1:
            colors[nodes] = -1
        else:
            if len(self.graph[nodes]) == 1 and parity % 2 == 0:
                if q == 1:
                    colors[nodes] = 1
                else:
                    colors[nodes] = -1
                    count += 1
            else:
                if parity % 2 == 0:
                    colors[nodes] = -1
                else:
                    colors[nodes] = 1
        visited_nodes.add(nodes)
        for neighbour in self.graph[nodes]:
            new_level = level + 1
            if neighbour not in visited_nodes:
                self.dfs_utility(neighbour, visited_nodes, colors, level - 1, new_level)

    def dfs(self, node):
        Visited = set()
        color = collections.defaultdict()
  ...