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.
//package problems;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class MaximumProduct {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(reader.readLine());
        while (t > 0) {
            int n = Integer.parseInt(reader.readLine());
            String temp[] = reader.readLine().split("\\s+");
            long[] nums = new long[n];
            int positives = 0, negatives = 0, zeros = 0;
            for (int i = 0; i < n; i++) {
                nums[i] = Integer.parseInt(temp[i]);
                if (nums[i] > 0)
                    positives++;
                else if (nums[i] < 0)
                    negatives--;
                else
                    zeros++;
            }

            if (n == 5) {
                long multiply = 1L;
                for (long item : nums)
                    multiply *= item;
                System.out.println(multiply);
            } else {
                if (n - zeros < 5) {
                    System.out.println(0);
                } else {
                    Arrays.sort(nums);
                    long multiplication = 1L;
                    int left = 0, right = n - 1, count = 0;
                    if (nums[left] * nums[left + 1] > nums[right] * nums[right - 1]) {
                        multiplication *= nums[left] * nums[left + 1];
                        count += 2;
                        left += 2;
                    } else {
                        multiplication *= nums[right] * nums[right - 1];
                        count += 2;
                        right -= 2;
                    }
                    if (multiplication * (nums[left] * nums[left + 1] * nums[right]) > multiplication * (nums[right] * nums[right - 1] * nums[right - 2]))
                        multiplication *= (nums[left] * nums[left + 1] * nums[right]);
                    else
                        multiplication *= (nums[right] * nums[right - 1] * nums[right - 2]);

                    multiplication = Math.max(multiplication, nums[0] * nums[1] * nums[2] * nums[3] * nums[n - 1]);
                    multiplication = Math.max(multiplication, nums[0] * nums[1] * nums[n - 1] * nums[n - 2] * nums[n - 3]);
                    multiplication = Math.max(multiplication, nums[n - 1] * nums[n - 2] * nums[n - 3] * nums[n - 4] * nums[n - 5]);
                    System.out.println(multiplication);
                }
            }
            t--;
        }

//        System.out.println(Long.MAX_VALUE);
    }
}
