#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10, NN = 2e3 + 10, INF = 0x3f3f3f3f, LEN = 20;
const long long MOD = 1e9 + 7;
const unsigned long long seed = 31;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
vector<int> vec;
int n, sum, temp;
int a[N];
bool vis[N];
void init() {
  vec.clear();
  memset(vis, false, sizeof vis);
}
int gcd(int x, int y) { return x == 0 ? y : gcd(y % x, x); }
bool cmp(int x, int y) { return x > y; }
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    init();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    sort(a + 1, a + 1 + n, cmp);
    sum = 1;
    temp = a[1];
    vis[1] = true;
    vec.push_back(a[1]);
    while (sum < n) {
      int id, max = -INF;
      for (int i = 1; i <= n; i++) {
        if (vis[i]) continue;
        int cnt = gcd(temp, a[i]);
        if (cnt > max) {
          id = i;
          max = cnt;
        }
      }
      ++sum;
      vis[id] = true;
      vec.push_back(a[id]);
      temp = gcd(temp, a[id]);
    }
    bool flag = false;
    vector<int>::iterator it;
    for (it = vec.begin(); it != vec.end(); it++) {
      if (!flag) {
        printf("%d", *it);
        flag = true;
      } else
        printf(" %d", *it);
    }
    puts("");
  }
}

### Prompt

Generate a cpp solution to 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;
const int N = 1e6 + 10, NN = 2e3 + 10, INF = 0x3f3f3f3f, LEN = 20;
const long long MOD = 1e9 + 7;
const unsigned long long seed = 31;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
vector<int> vec;
int n, sum, temp;
int a[N];
bool vis[N];
void init() {
  vec.clear();
  memset(vis, false, sizeof vis);
}
int gcd(int x, int y) { return x == 0 ? y : gcd(y % x, x); }
bool cmp(int x, int y) { return x > y; }
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    init();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    sort(a + 1, a + 1 + n, cmp);
    sum = 1;
    temp = a[1];
    vis[1] = true;
    vec.push_back(a[1]);
    while (sum < n) {
      int id, max = -INF;
      for (int i = 1; i <= n; i++) {
        if (vis[i]) continue;
        int cnt = gcd(temp, a[i]);
        if (cnt > max) {
          id = i;
          max = cnt;
        }
      }
      ++sum;
      vis[id] = true;
      vec.push_back(a[id]);
      temp = gcd(temp, a[id]);
    }
    bool flag = false;
    vector<int>::iterator it;
    for (it = vec.begin(); it != vec.end(); it++) {
      if (!flag) {
        printf("%d", *it);
        flag = true;
      } else
        printf(" %d", *it);
    }
    puts("");
  }
}

```