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;
int n;
const int maxn = 1e5 + 10;
long long dp[maxn][6][2];
long long a[maxn];
const long long inf = 1e18;
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      for (int j = 1; j <= 5; j++) dp[i][j][0] = inf, dp[i][j][1] = -inf;
    }
    dp[1][1][0] = dp[1][1][1] = a[1];
    for (int i = 2; i <= n; i++) {
      dp[i][1][0] = min(a[i], dp[i - 1][1][0]);
      dp[i][1][1] = max(a[i], dp[i - 1][1][1]);
      for (int j = 2; j <= min(i, 5); j++) {
        if (a[i] >= 0) {
          dp[i][j][0] = min(dp[i - 1][j][0], dp[i - 1][j - 1][0] * a[i]);
          dp[i][j][1] = max(dp[i - 1][j][1], dp[i - 1][j - 1][1] * a[i]);
        } else {
          dp[i][j][0] = min(dp[i - 1][j][0], dp[i - 1][j - 1][1] * a[i]);
          dp[i][j][1] = max(dp[i - 1][j][1], dp[i - 1][j - 1][0] * a[i]);
        }
      }
    }
    cout << dp[n][5][1] << endl;
  }
  return 0;
}
