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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main
{
    public static void main(String[] args)
    {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB
    {
        public void solve(int testNumber, Scanner in, PrintWriter out)
        {
            int T = in.nextInt();
            while (T-- > 0)
            {
                solveOne(in, out);
            }
        }

        private void solveOne(Scanner in, PrintWriter out)
        {
            int N = in.nextInt();
            int nums[] = CPUtils.readIntArray(N, in);

            if (nums.length == 5)
            {
                long p = 1;
                for (int idx = 0; idx < 5; idx++)
                {
                    p *= nums[idx];
                }
                out.println(p);
                return;
            }

            Arrays.sort(nums);

            if (nums[nums.length - 1] < 0)
            {
                long p = 1;
                for (int idx = nums.length - 1; idx >= nums.length - 5; idx--)
                {
                    p *= nums[idx];
                }
                out.println(p);
                return;
            }

            if (nums[nums.length - 1] == 0)
            {
                out.println(0);
                return;
            }


            long pMax = nums[nums.length - 1];
            int i = 0;
            int j = nums.length - 2;
            int miss = 4;
            while (miss > 0)
            {
                if ((nums[i] * nums[i + 1]) >= (nums[j] * nums[j - 1]))
                {
                    pMax *= nums[i] * nums[i + 1];
                    i += 2;
                } else
                {
                    pMax *= nums[j] * nums[j - 1];
                    j -= 2;
                }
                miss -= 2;
            }
            out.println(pMax);
        }

    }

    static class CPUtils
    {
        public static int[] readIntArray(int size, Scanner in)
        {
            int[] array = new int[size];
            for (int i = 0; i < size; i++)
            {
                array[i] = in.nextInt();
            }
            return array;
        }

    }
}

