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;
long long m, n;
long long ar[200009];
int main() {
  long long i, j, test = 1, cas = 0;
  long long a, b, mx, p;
  scanf("%lld", &test);
  while (test--) {
    scanf("%lld", &n);
    mx = -1;
    for (i = 0; i < n; i++) {
      scanf("%lld", &ar[i]);
      if (ar[i] == 0) mx = 0;
    }
    sort(ar, ar + n);
    if (mx == -1) {
      mx = 1;
      for (i = 0; i < 5; i++) {
        mx *= ar[i];
      }
    }
    for (i = 0; i <= 5; i++) {
      p = 1;
      for (j = 0; j < i; j++) {
        p *= ar[j];
      }
      for (j = 0; j < (5 - i); j++) {
        p *= ar[n - 1 - j];
      }
      mx = max(mx, p);
    }
    printf("%lld\n", mx);
  }
  return 0;
}
