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.util.*;

public class A {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        StringBuilder sb = new StringBuilder();
        int testCases = sc.nextInt();
        for (int t = 0; t < testCases; t++) {
            int n = sc.nextInt();
            int arr[] = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = sc.nextInt();
            }
            Arrays.sort(arr);
            // Back 0 front 5
            long ans1 = 1;
            int count = 0;
            while (count != 5) {
                ans1 *= arr[n - count - 1];
                count++;
            }
            // b2 f3
            count = 0;
            long ans2 = 1;
            while (count != 2) {
                ans2 *= arr[count];
                count++;
            }
            count = 0;
            while (count != 3) {
                ans2 *= arr[n - count - 1];
                count++;
            }

            // b4 front1
            long ans3 = 1;
            count = 0;
            while (count != 4) {
                ans3 *= arr[count];
                count++;
            }
            ans3 *= arr[n - 1];

            long max = Math.max(ans1, Math.max(ans2, ans3));

            sb.append(max).append("\n");
        }
        System.out.print(sb);
    }

}