#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a == 0) return 0;
  if (b == 0) return 0;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
int main() {
  long long t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long> v(n);
    for (long long i = 0; i < n; i++) cin >> v[i];
    long long idx = max_element(v.begin(), v.end()) - v.begin();
    long long prevgcd = v[idx];
    v[idx] = 0;
    cout << prevgcd << " ";
    for (int i = 0; i < n - 1; i++) {
      long long m = 0;
      for (int j = 0; j < n; j++) {
        long long temp = gcd(prevgcd, v[j]);
        if (m < temp) {
          m = temp;
          idx = j;
        }
      }
      cout << v[idx] << ' ';
      prevgcd = m;
      v[idx] = 0;
    }
    cout << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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, 1].

In the third test case of the example, number 9 should be the first in b, and GCD(9, 3)=3, GCD(9, 8)=1, so the second number of b should be 3.

In the seventh test case of the example, first four numbers pairwise have a common divisor (a power of two), but none of them can be the first in the optimal permutation b.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a == 0) return 0;
  if (b == 0) return 0;
  if (a == b) return a;
  if (a > b) return gcd(a - b, b);
  return gcd(a, b - a);
}
int main() {
  long long t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long> v(n);
    for (long long i = 0; i < n; i++) cin >> v[i];
    long long idx = max_element(v.begin(), v.end()) - v.begin();
    long long prevgcd = v[idx];
    v[idx] = 0;
    cout << prevgcd << " ";
    for (int i = 0; i < n - 1; i++) {
      long long m = 0;
      for (int j = 0; j < n; j++) {
        long long temp = gcd(prevgcd, v[j]);
        if (m < temp) {
          m = temp;
          idx = j;
        }
      }
      cout << v[idx] << ' ';
      prevgcd = m;
      v[idx] = 0;
    }
    cout << endl;
  }
  return 0;
}

```