#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.14159265358979323846;
long long MOD = 1e9 + 7;
const char nl = '\n';
const long long inf = 1e15;
long long nxt() {
  long long x;
  cin >> x;
  return x;
}
long long mul(long long x, long long y) {
  return (1ll * (x % MOD) * (y % MOD));
}
long long modpow(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2) z = mul(z, x);
    x = mul(x, x);
    y /= 2;
  }
  return z;
}
long long power(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2) z = z * x;
    x = x * x;
    y /= 2;
  }
  return z;
}
long long gcd(long long a, long long b) {
  if (a < b) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long sq(long long a) {
  long long ans = (1ll * a * a);
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  vector<long long> c(1001);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    c[a[i]]++;
  }
  sort(a, a + n);
  reverse(a, a + n);
  long long b[n];
  b[0] = a[0];
  long long gc = a[0];
  c[a[0]]--;
  cout << a[0] << " ";
  for (long long i = (1); i <= n - 1; i++) {
    long long mx = 1;
    long long pos = -1;
    for (long long j = 1; j < n; j++) {
      if (c[a[j]]) {
        if (a[j] < mx) break;
        long long k = gcd(a[j], gc);
        if (k > mx) {
          mx = k;
          pos = j;
        }
      }
    }
    if (pos == -1) {
      for (long long i = 0; i < n; i++) {
        if (c[a[i]]) {
          cout << a[i] << " ";
          c[a[i]]--;
        }
      }
      break;
    }
    cout << a[pos] << " ";
    c[a[pos]]--;
    gc = mx;
  }
  cout << nl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long TC = 1;
  cin >> TC;
  for (long long TT = 0; TT < TC; TT++) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

### Prompt

Your task is to create 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 long double pi = 3.14159265358979323846;
long long MOD = 1e9 + 7;
const char nl = '\n';
const long long inf = 1e15;
long long nxt() {
  long long x;
  cin >> x;
  return x;
}
long long mul(long long x, long long y) {
  return (1ll * (x % MOD) * (y % MOD));
}
long long modpow(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2) z = mul(z, x);
    x = mul(x, x);
    y /= 2;
  }
  return z;
}
long long power(long long x, long long y) {
  long long z = 1;
  while (y > 0) {
    if (y % 2) z = z * x;
    x = x * x;
    y /= 2;
  }
  return z;
}
long long gcd(long long a, long long b) {
  if (a < b) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long sq(long long a) {
  long long ans = (1ll * a * a);
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  vector<long long> c(1001);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    c[a[i]]++;
  }
  sort(a, a + n);
  reverse(a, a + n);
  long long b[n];
  b[0] = a[0];
  long long gc = a[0];
  c[a[0]]--;
  cout << a[0] << " ";
  for (long long i = (1); i <= n - 1; i++) {
    long long mx = 1;
    long long pos = -1;
    for (long long j = 1; j < n; j++) {
      if (c[a[j]]) {
        if (a[j] < mx) break;
        long long k = gcd(a[j], gc);
        if (k > mx) {
          mx = k;
          pos = j;
        }
      }
    }
    if (pos == -1) {
      for (long long i = 0; i < n; i++) {
        if (c[a[i]]) {
          cout << a[i] << " ";
          c[a[i]]--;
        }
      }
      break;
    }
    cout << a[pos] << " ";
    c[a[pos]]--;
    gc = mx;
  }
  cout << nl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long TC = 1;
  cin >> TC;
  for (long long TT = 0; TT < TC; TT++) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}

```