You are given an array of integers a_1,a_2,…,a_n. Find the maximum possible value of a_ia_ja_ka_la_t among all five indices (i, j, k, l, t) (i<j<k<l<t).

Input

The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (5≤ n≤ 10^5) — the size of the array.

The second line of each test case contains n integers a_1,a_2,…,a_n (-3× 10^3≤ a_i≤ 3× 10^3) — given array.

It's guaranteed that the sum of n over all test cases does not exceed 2⋅ 10^5.

Output

For each test case, print one integer — the answer to the problem.

Example

Input


4
5
-1 -2 -3 -4 -5
6
-1 -2 -3 1 2 -1
6
-1 0 0 0 -1 -1
6
-9 -7 -5 -3 -2 1


Output


-120
12
0
945

Note

In the first test case, choosing a_1,a_2,a_3,a_4,a_5 is a best choice: (-1)⋅ (-2) ⋅ (-3)⋅ (-4)⋅ (-5)=-120.

In the second test case, choosing a_1,a_2,a_3,a_5,a_6 is a best choice: (-1)⋅ (-2) ⋅ (-3)⋅ 2⋅ (-1)=12.

In the third test case, choosing a_1,a_2,a_3,a_4,a_5 is a best choice: (-1)⋅ 0⋅ 0⋅ 0⋅ (-1)=0.

In the fourth test case, choosing a_1,a_2,a_3,a_4,a_6 is a best choice: (-9)⋅ (-7) ⋅ (-5)⋅ (-3)⋅ 1=945.
#include <bits/stdc++.h>
using namespace std;
int64_t solve(vector<int> arr, int n) {
  int64_t pr = 1;
  if (n == 5) {
    for (int i = 0; i < n; i++) {
      pr = pr * 1ll * arr[i];
    }
    return pr;
  }
  sort(arr.begin(), arr.end());
  int cnt0 = 0;
  for (int i = n - 5; i <= n - 1; i++) {
    if (arr[i] != 0)
      pr = pr * 1ll * arr[i];
    else
      cnt0++;
  }
  int64_t mx = (cnt0 == 0) ? pr : 0;
  for (int i = 0; i < 5; i++) {
    pr = (arr[n - 5 + i] != 0) ? pr / arr[(n - 5 + i)] : pr;
    if (arr[n - 5 + i] == 0) cnt0--;
    pr = pr * 1ll * arr[i];
    if (cnt0 > 0)
      mx = 0;
    else if (mx < pr)
      mx = pr;
  }
  return mx;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    cout << solve(arr, n) << endl;
  }
  return 0;
}
