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.
//package September;

import java.io.*;
import java.util.Arrays;

public class B1406 {
    public static void main(String[] args) throws IOException {
        BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));

        int test = Integer.parseInt(scan.readLine().trim());
        String line[];
        test:
        while (test-- > 0) {
            int n = Integer.parseInt(scan.readLine().trim());
            line = scan.readLine().split("\\s+");

            long arr[] = new long[n];
            for (int i = 0; i < n; i++) {
                arr[i] = Long.parseLong(line[i]);
            }

            Arrays.sort(arr);
            long ans = 1;

            for(int i = 0; i < 5; i++) {
                ans *= arr[n - i - 1];
            }
            if(arr[n-1]>=0){
                long val=arr[0]*arr[1]*arr[2]*arr[3]*arr[n-1];
                ans=Math.max(ans,val);
            }
            if(arr[n-3]>=0)
            {	long val=arr[0]*arr[1]*arr[n-1]*arr[n-2]*arr[n-3];
                ans=Math.max(ans,val);
            }
//            cout<<ans<<endl;
            System.out.println(ans);
        }
    }
}
