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 jhgjhh {
    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        PrintStream out = new PrintStream(System.out);
        int n = in.nextInt();
        for (int i = 0; i<n; i++){
            int k = in.nextInt();
            int [] mas = new int[k];
            for(int j = 0; j<k;j++){
                mas[j]= in.nextInt();
            }
            Arrays.sort(mas);

            /*int i1 = mas[k - 2] * mas[k - 3];
            if((mas[0]*mas[1])> i1){
                ans = mas[0]*mas[1]*ans;
                if(mas[2]*mas[3]>mas[k-2]*mas[k-3]){
                    ans = ans * mas[2]*mas[3];
                }
                else{
                    ans = ans * mas[k-2]*mas[k-3];
                }
            }
            else{
                ans = mas[k-2]*mas[k-3]*ans;
                if(mas[0]*mas[1]>mas[k-4]*mas[k-5]){
                    ans = mas[0]*mas[1]*ans;
                }
                else{
                    ans = ans * mas[k-4]*mas[k-5];
                }
            }*/
            long ans = mas[k-1];
            if (mas[k - 1] < 0) {
                ans = ans* mas[k-2]*mas[k-3]*mas[k-4]*mas[k-5];
            }
            else{
                ans = ans * Math.max(mas[0]*mas[1], mas[k-4]*mas[k-5])*Math.max(mas[2]*mas[3],mas[k-2]*mas[k-3]);

            }
            out.println(ans);
        }
    }
}