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 Main {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(in.readLine());
        while (t-- > 0) {
            int n = Integer.parseInt(in.readLine());
            String[] inp2 = in.readLine().split("\\s+");
            ArrayList<Long> arr = new ArrayList<>();
            for (int i=0;i<n;i++) {
                long e = Long.parseLong(inp2[i]);
                arr.add(e);
            }
            Collections.sort(arr);
            long maxprod = 1;
            if(n==5){
                maxprod = arr.get(0)*arr.get(1)*arr.get(2)*arr.get(3)*arr.get(4);
                System.out.println(maxprod);
                continue;
            }

            long ans1 = arr.get(n-1)*arr.get(0)*arr.get(1)*arr.get(2)*arr.get(3);
            long ans2 = arr.get(n-1)*arr.get(n-2)*arr.get(n-3)*arr.get(0)*arr.get(1);
            long ans3 = arr.get(0)*arr.get(1)*arr.get(2)*arr.get(3)*arr.get(4);
            maxprod = arr.get(n-1)*arr.get(n-2)*arr.get(n-3)*arr.get(n-4)*arr.get(n-5);
            maxprod = (ans1>maxprod)?ans1:maxprod;
            maxprod = (ans2>maxprod)?ans2:maxprod;
            maxprod = (ans3>maxprod)?ans3:maxprod;
            System.out.println(maxprod);
        }
    }
}
