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 C670_2 {
	static long ans = Long.MIN_VALUE;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while (t > 0) {
			t--;
			ans = Long.MIN_VALUE;
			int n = sc.nextInt();
			int arr[] = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = sc.nextInt();
			}
			Arrays.sort(arr);
			int size = n < 10 ? n : 10;
			long ar[] = new long[size];
			for (int i = 0; i < 5; i++) {
				ar[i] = arr[i];
			}
			int i = 5, j = n-1;
			while (i < size) {
				ar[i] = arr[j];
				j--; i++;
			}
			find(ar, size, 0, 1, 5);
			
			System.out.println(ans);
		}

	}
	static void find(long arr[], int n, int cur, long mul,  int temp) {
		if (cur == n) {
			if (temp == 0) {
				ans = Math.max(mul, ans);
			}
			return;
		}
		find(arr, n, cur + 1, mul * arr[cur], temp - 1);
		find(arr, n, cur + 1, mul, temp);
		
	}

}
