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 zad {


    public static void main(String[] s) throws IOException {
    Scanner scanner=new Scanner(System.in);
    long t =scanner.nextInt();
    int n;
    long m;
    while(t-->0){
        List<Long> list=new ArrayList<>();
        n =scanner.nextInt();
        for(int i=0;i<n;i++)
        list.add(scanner.nextLong());
        Collections.sort(list);
        m=Collections.max(list);
        if(m<0) {
            System.out.println(list.get(n-5) * list.get(n-4) * list.get(n-3) * list.get(n-2) * list.get(n-1));
            continue;
        }


        long first=list.get(n-5) * list.get(n-4) * list.get(n-3) * list.get(n-2) * list.get(n-1);
        long second=list.get(0) * list.get(1) * list.get(n-3) * list.get(n-2) * list.get(n-1);
        long third=list.get(0) * list.get(1) * list.get(2) * list.get(3) * list.get(n-1);
        long end=Long.max(first,second);
        end=Long.max(end,third);
        System.out.println(end);
    }
    }
}
