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

public class Main1 {

    public static void main(String[] args) throws IOException {
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int t = Integer.parseInt(st.nextToken());

        while(t-- != 0){
            int n = Integer.parseInt(new StringTokenizer(br.readLine()).nextToken());
            ArrayList<Integer> arr = new ArrayList<>();

            st = new StringTokenizer(br.readLine());
            for(int i = 0; i < n; i++){
                arr.add(Integer.parseInt(st.nextToken()));
            }
            Collections.sort(arr);

            long maximum = Long.MIN_VALUE;
            for(int len = 0; len <= 5; len++){
                long temporalMaximum = 1;
                // infront
                for(int i = 0; i < len; i++){
                    temporalMaximum *= arr.get(i);
                }
                for(int i = 0; i < 5-len; i++){
                    temporalMaximum*=arr.get(arr.size()-1-i);
                }
                maximum = Math.max(maximum, temporalMaximum);
            }
            bw.write(maximum+"\n");
            bw.flush();
        }
    }
}
