human
What is 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,...
assistant
#Here is Python code to fix of b, but a ≠ b;
.

# import math
# from collections import Counter
#from collections import defaultdict
from collections import deque
# from bisect import bisect_left as bl, bisect_right as br
# from itertools import accumulate,permutations as perm,combinations as comb
# from heapq import heapify,heappush,heappop,nlargest,nsmallest
# from fractions import Fraction
import sys
from functools import reduce  # , lru_cache
"""author-bravotango. FastIO code by pajenegod"""

# help(deque)

def factors(n):
    return sorted(set(reduce
                      (list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))), reverse=True)

def main():
    t = Iint()
    for _ in range(t):
        n = Iint()
        a = Ilist()
        a.sort(reverse=True)
        b = [0]*n
        b[0] = max(a)
        a[0] = -1
        prev = b[0]
        for i in range(1, n):
            debug(b, c=1)
            f = factors(prev)
            debug(f, c=3)
            for fi in f:
                debug(a, fi)
                for j in range(n):
                    if a[j] != -1 and a[j] % fi == 0:
                        b[i] = a[j]
                        a[j] = -1
                        prev = fi
                        break
                else:
                    continue
                break
        Plist(b)

en = enumerate
# IO region1
FIO = 10

def debug(*args, c=6): pass  # print('\033[3{}m'.format(c), *args, '\033[0m', file=sys.stderr)

BUFSIZE = 8*1024

def Istr(): return input()

def Iint(): return int(input())

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

def Imap(): return map(int, input().split())   # int map

def Plist(li, s=' '): print(s.join(map(str, li)))   # non string iterable

if not FIO:
    if __name__ == '__main__':
        main()
        exit()
else:
    from io import BytesIO, IOBase
    import os
# FastIO

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode...