human
Give me a Python code to fix of b, but a ≠ b; given the following:
Alexander is a well-known programmer. Today he decided to finally go out and play football, but with the first hit he left a dent on the new Rolls-Royce of the wealthy businessman Big Vova. Vladimir has recently opened a store on the popular online marketplace "Zmey-Gorynych", and offers Alex a job: if he shows his programming skills by solving a task, he'll work as a cybersecurity specialist. Otherwise, he'll be delivering some doubtful products for the next two years.

You're given n positive integers a_1, a_2, ..., a_n. Using each of them exactly at once, you're to make such sequence b_1, b_2, ..., b_n that sequence c_1, c_2, ..., c_n is lexicographically maximal, where c_i=GCD(b_1,...,b_i) - the greatest common divisor of the first i elements of b. 

Alexander is really afraid of the conditions of this simple task, so he asks you to solve it.

A sequence a is lexicographically smaller than a sequence b if and only if one of the following holds:

  * a is a prefix of b, but a ≠ b;
  * in the first position where a and b differ, the sequence a has a smaller element than the corresponding element in b.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^3). Description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^3) — the length of the sequence a.

The second line of each test case contains n integers a_1,...,a_n (1 ≤ a_i ≤ 10^3) — the sequence a.

It is guaranteed that the sum of n over all test cases does not exceed 10^3.

Output

For each test case output the answer in a single line — the desired sequence b. If there are multiple answers, print any.

Example

Input

7
2
2 5
4
1 8 2 3
3
3 8 9
5
64 25 75 100 50
1
42
6
96 128 88 80 52 7
5
2 4 8 16 17

Output

5 2 
8 2 1 3 
9 3 8 
100 50 25 75 64 
42 
128 96 80 88 52 7 
17 2 4 8 16 

Note

In the first test case of the example, there are only two possible permutations b — [2, 5] and [5, 2]: for the first one c=[2, 1], for the second one c=[5,...
Using python3 can you solve the prior task?
assistant
#Here is Python code to fix of b, but a ≠ b;
.

from __future__ import division, print_function

import os
import sys
from io import BytesIO, IOBase

def main():
    def solve():
        from math import inf, gcd

        def removeFromNumbersLeft(number):
            if numbersLeft[number] == 1:
                del numbersLeft[number]
            else:
                numbersLeft[number] -= 1

        n = int(input())
        numbersLeft = {}
        for num in map(lambda x: int(x), input().split()):
            if num in numbersLeft:
                numbersLeft[num] += 1
            else:
                numbersLeft[num] = 1

        b = [0] * n
        b[0] = max(numbersLeft)
        removeFromNumbersLeft(b[0])

        previousGCD = b[0]

        for i in range(1, n):
            maxGCD = -inf

            for num in numbersLeft:
                newGCD = gcd(previousGCD, num)
                if newGCD > maxGCD:
                    maxGCD = newGCD
                    bestNum = num

            b[i] = bestNum
            removeFromNumbersLeft(bestNum)
            previousGCD = maxGCD

        print(' '.join(map(lambda x: str(x), b)))

    t = int(input())
    for i in range(t):
        solve()

BUFFSIZE = 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, BUFFSIZE))
            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, BUFFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
           ...