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>
const char nl = '\n';
using namespace std;
class Input {
 public:
  Input() {
    myPc();
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
  }
  void myPc() {}
};
Input setup;
void solve(long long int tc) {
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long int ans;
  ans = a[0] * a[1] * a[2] * a[3] * a[4];
  ans = max(ans, a[0] * a[1] * a[n - 1] * a[n - 2] * a[n - 3]);
  ans = max(ans, a[0] * a[1] * a[2] * a[3] * a[n - 1]);
  ans = max(ans, a[n - 1] * a[n - 2] * a[n - 3] * a[n - 4] * a[n - 5]);
  cout << ans;
}
int32_t main() {
  long long int tcc;
  tcc = 1;
  cin >> tcc;
  for (long long int i = 1; i <= tcc; i++) {
    solve(i);
    cout << "\n";
  }
}
