构造出一个数列,使得它的前缀 $GCD$ 的字典序最大,数组范围只有 $1000$,所以在构造的过程中,计算当前的前缀 $GCD$,线性选出一个和之前结果最大的 $GCD(pre, a[i])$ 即可。

// Date: Tue Dec  5 23:51:39 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 1010;
int t, n, a[N], b[N];
bool vis[N];

int gcd(int a, int b) { return (a % b == 0) ? b : gcd(b, a % b); }

int main(void) {
#ifdef _DEBUG
  freopen("1407b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    memset(vis, false, sizeof vis);
    For1(i, 1, n) { cin >> a[i]; }

    sort(a + 1, a + 1 + n, greater<int>());
    b[1] = a[1];
    vis[1] = true;
    int pre = b[1];

    For1(i, 2, n) {
      int g = 0, idx = -1;
      For1(j, 2, n) {
        if (vis[j])
          continue;
        int g1 = gcd(pre, a[j]);
        if (g1 > g) {
          g = g1;
          idx = j;
        }
      }
      vis[idx] = true;
      b[i] = a[idx];
      pre = gcd(pre, b[i]);
    }

    For1(i, 1, n) { cout << b[i] << ' '; }
    cout << '\n';
  }

  return 0;
}