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.Arrays;
import java.util.Scanner;

public class Main {

    public static void main(String args[]) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while (t-- > 0) {
            int n = scanner.nextInt();
            long[] a = new long[n];
            int noOfNegs = 0;
            int noOfPos = 0;
            int noOfZeros = 0;
            for (int i = 0; i < n; i++) {
                a[i] = scanner.nextInt();
                if (a[i] > 0) {
                    noOfPos++;
                } else if (a[i] < 0) {
                    noOfNegs++;
                } else {
                    noOfZeros++;
                }
            }

            Arrays.sort(a);

            long ans = Long.MIN_VALUE;
            if (noOfPos >= 5) {
                // 0 -ves , 5 +ves
                long x = a[n - 1] * a[n - 2] * a[n - 3] * a[n - 4] * a[n - 5];
                ans = Math.max(ans,x);
            }
            if (noOfPos >=3 && noOfNegs >=2) {
                // 2 -ves , 3 +ves
                long y = a[0] * a[1] * a[n - 1] * a[n - 2] * a[n - 3];
                ans = Math.max(ans,y);
            }
            if (noOfNegs >=4 && noOfPos >=1) {
                // 4 -ves , 1 +ves
                long z = a[0] * a[1] * a[2] * a[3] * a[n - 1];
                ans = Math.max(ans,z);
            }
            if (noOfZeros >= 1) {
                // At lease one zero is present
                ans = Math.max(ans, 0L);
            }
            if (noOfNegs >=5 && noOfZeros == 0 && noOfPos == 0) {
                // 5 -ves , 0 +ves
                long m = a[n - 1] * a[n - 2] * a[n - 3] * a[n - 4] * a[n - 5];
                ans = Math.max(ans, m);
            }
            if (n == 5) {
                // Only 5 nos given
                long m = a[0] * a[1] * a[2] * a[3] * a[4];
                ans = Math.max(ans, m);
            }
            System.out.println(ans);
        }
    }
}
