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;
const long long mod = 1e9 + 7;
long long find(long long a[], long long n) {
  if (n < 5) return -1;
  sort(a, a + n);
  long long last = a[n - 1] * a[n - 2] * a[n - 3] * a[n - 4] * a[n - 5];
  long long first = a[0] * a[1] * a[2] * a[3] * a[4];
  long long two = a[0] * a[1] * a[n - 1] * a[n - 2];
  if (two * a[2] > (two * a[n - 3]))
    two *= a[2];
  else
    two = a[n - 3];
  return max(last, max(first, two));
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n], i;
    for (i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    long long last = a[n - 1] * a[n - 2] * a[n - 3] * a[n - 4] * a[n - 5];
    long long first = a[0] * a[1] * a[2] * a[3] * a[4];
    long long two = a[0] * a[1] * a[n - 1] * a[n - 2];
    if ((two * a[2]) > (two * a[n - 3]))
      two *= a[2];
    else
      two *= a[n - 3];
    long long four = max(a[0] * a[n - 1] * a[n - 2] * a[n - 3] * a[n - 4],
                         a[0] * a[1] * a[2] * a[3] * a[n - 1]);
    cout << max({last, first, two, four}) << "\n";
  }
  return 0;
}
