human
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,...
The above is tricky. Write me a correct solution in PYTHON3.
 Write me a Python code to fix of b, but a ≠ b;?
assistant
#Here is Python code to fix of b, but a ≠ b;
.

# It's never too late to start!
from bisect import bisect_left
import os
import sys
from io import BytesIO, IOBase
from collections import Counter, defaultdict
from collections import deque
from functools import cmp_to_key
import math
import heapq

def sin():
    return input()
def ain():
    return list(map(int, sin().split()))
def sain():
    return input().split()
def iin():
    return int(sin())
MAX = float('inf')
MIN = float('-inf')
MOD = 1000000007
def sieve(n): 
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    s = set()
    for p in range(2, n+1): 
        if prime[p]: 
            s.add(p)
    return s
def readTree(n, m):
    adj = [deque([]) for _ in range(n+1)]
    for _ in range(m):
        u,v = ain()
        adj[u].append(v)
        adj[v].append(u)
    return adj

# Stay hungry, stay foolish!

def main():
    for _ in range(iin()):
        n = iin()
        a = ain()
        ans = deque([])
        v = [0]*(n)
        maxi = max(a)
        idx = a.index(maxi)
        ans.append(maxi)
        v[idx] = 1
        c = n-1
        curr_gcd = maxi
        while c>0:
            d = deque([])
            maxi = -1
            for i in range(n):
                if v[i] == 0:
                    x = math.gcd(curr_gcd, a[i])
                    if x>maxi:
                        maxi = x
                        d = deque([i])
                    elif x == maxi:
                        d.append(i)
            for j in d:
                c -= 1
                v[j] = 1
                ans.append(a[j])
            curr_gcd = maxi
        print(*ans)


        
















    



            
        

        









    
    


# Fast IO Template starts

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in...