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 maximum_product {

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner s = new Scanner(System.in);
		
		int t = s.nextInt();
		
		while(t-->0) {
			
			int n= s.nextInt();
			int[] arr = new int[n];
			for(int i=0; i<n; i++) {
				arr[i] = s.nextInt();
			}
			
			Arrays.sort(arr);
			
			long p1 = 1;
			long p2 = 1;
			long p3 = 1;
			long p4 = 1;
			
			for(int i=arr.length-5 ; i<arr.length;i++) {
				p1 = 1l*p1*arr[i];
			}
			
			
			
			for(int i=arr.length-3 ; i<arr.length;i++) {
				p2 = 1l*p2*arr[i];
			}
			p2 = 1l* p2 *arr[0]*arr[1];
			

			
			for(int i=arr.length-1 ; i<arr.length;i++) {
				p3 = 1l*p3*arr[i];
			}
			p3 = 1l*p3*arr[0]*arr[1]*arr[2]*arr[3];
			
	
			for(int i=0 ; i<5;i++) {
				p4 = 1l*p4*arr[i];
			}
	
			System.out.println(Math.max(p1, Math.max(p2, Math.max(p3, p4))));
		}
	}

}
