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.BufferedInputStream;
import java.io.IOException;
import java.util.Arrays;
import java.util.Scanner;

/**
 * Created by Harry on 5/31/20.
 */
public class Test {
    public static void main(String[] args) throws IOException {
        Scanner scanner = new Scanner(new BufferedInputStream(System.in));
        int T = scanner.nextInt();
        for(int t=1; t<=T; t++){
            int n = scanner.nextInt();
            long[] nums = new long[n];
            for(int i=0; i<n; i++){
                nums[i] = scanner.nextLong();
            }
            long res = Long.MIN_VALUE;
            Arrays.sort(nums);
            for(int i=0; i<5; i++){
                long cur = 1;
                for(int j=0; j<i; j++){
                    cur *= nums[j];
                }
                for(int j=n-5+i; j<n; j++){
                    cur *= nums[j];
                }
                res = Math.max(res, cur);
            }
            System.out.println(res);
        }
    }
}
