#include <bits/stdc++.h>
using namespace std;
void speed() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
}
int gcd(int a, int b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
int lcm(int a, int b) { return (a / gcd(a, b)) * b; }
long long int PowerOf2Num(long long int n) { return (n & (~(n - 1))); }
bool isPow(long long int x) { return x && (!(x & (x - 1))); }
bool divByDig(long long int num) {
  long long int x = num;
  while (x > 0) {
    long long int y = x % 10;
    x /= 10;
    if (y != 0 && num % y != 0) {
      return false;
    }
  }
  return true;
}
long long int numDig(long long int num) {
  long long int dig = 0;
  while (num > 0) {
    dig++;
    num /= 10;
  }
  return dig;
}
bool sameDig(long long int num) {
  long long int x = num;
  long long int same = x % 10;
  x /= 10;
  while (x > 0) {
    long long int yo = x % 10;
    if (same != yo) {
      return false;
    }
    x /= 10;
  }
  return true;
}
string decToBin(long long int n) {
  string s = "";
  for (long long int i = 63; i >= 0; i--) {
    long long int k = n >> i;
    if (k & 1) {
      s += "1";
    } else {
      s += "0";
    }
  }
  return s;
}
bool isPrime(long long int n) {
  if (n <= 1) {
    return false;
  }
  if (n <= 3) {
    return true;
  }
  if (n % 2 == 0 || n % 3 == 0) {
    return false;
  }
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int helper(vector<int> &v, int l, int r, int key) {
  while (r - l > 1) {
    int m = l + (r - l) / 2;
    if (v[m] >= key) {
      r = m;
    } else {
      l = m;
    }
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    map<int, int> mps;
    int maxi = 0;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      mps[x]++;
      maxi = max(maxi, x);
    }
    cout << maxi << " ";
    mps[maxi]--;
    int i = 1;
    while (i < n) {
      int ele = -1;
      int yo = 0;
      for (auto it : mps) {
        if (it.second > 0) {
          int x = it.first;
          int g = gcd(x, maxi);
          if (g > yo) {
            ele = x;
            yo = g;
          }
        }
      }
      maxi = yo;
      i += mps[ele];
      while (mps[ele] > 0) {
        cout << ele << " ";
        mps[ele]--;
      }
    }
    cout << '\n';
  }
}

### 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;
void speed() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
}
int gcd(int a, int b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
int lcm(int a, int b) { return (a / gcd(a, b)) * b; }
long long int PowerOf2Num(long long int n) { return (n & (~(n - 1))); }
bool isPow(long long int x) { return x && (!(x & (x - 1))); }
bool divByDig(long long int num) {
  long long int x = num;
  while (x > 0) {
    long long int y = x % 10;
    x /= 10;
    if (y != 0 && num % y != 0) {
      return false;
    }
  }
  return true;
}
long long int numDig(long long int num) {
  long long int dig = 0;
  while (num > 0) {
    dig++;
    num /= 10;
  }
  return dig;
}
bool sameDig(long long int num) {
  long long int x = num;
  long long int same = x % 10;
  x /= 10;
  while (x > 0) {
    long long int yo = x % 10;
    if (same != yo) {
      return false;
    }
    x /= 10;
  }
  return true;
}
string decToBin(long long int n) {
  string s = "";
  for (long long int i = 63; i >= 0; i--) {
    long long int k = n >> i;
    if (k & 1) {
      s += "1";
    } else {
      s += "0";
    }
  }
  return s;
}
bool isPrime(long long int n) {
  if (n <= 1) {
    return false;
  }
  if (n <= 3) {
    return true;
  }
  if (n % 2 == 0 || n % 3 == 0) {
    return false;
  }
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int helper(vector<int> &v, int l, int r, int key) {
  while (r - l > 1) {
    int m = l + (r - l) / 2;
    if (v[m] >= key) {
      r = m;
    } else {
      l = m;
    }
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    map<int, int> mps;
    int maxi = 0;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      mps[x]++;
      maxi = max(maxi, x);
    }
    cout << maxi << " ";
    mps[maxi]--;
    int i = 1;
    while (i < n) {
      int ele = -1;
      int yo = 0;
      for (auto it : mps) {
        if (it.second > 0) {
          int x = it.first;
          int g = gcd(x, maxi);
          if (g > yo) {
            ele = x;
            yo = g;
          }
        }
      }
      maxi = yo;
      i += mps[ele];
      while (mps[ele] > 0) {
        cout << ele << " ";
        mps[ele]--;
      }
    }
    cout << '\n';
  }
}

```