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.*;
import java.util.*;


public class MaximumP
{
	public static void main(String[]args)
	{
		Scanner sc = new Scanner(System.in);
		int t;
		t = sc.nextInt();
		while(t!=0)
		{
			t--;
			int n;
			n = sc.nextInt();
			long ar[] = new long[n];
			for(int i=0;i<n;i++)
			{
				ar[i] = sc.nextInt();
				
			}
			if(n==5)
			{
				System.out.println(ar[0]*ar[1]*ar[2]*ar[3]*ar[4]);
			}
			else{
			Arrays.sort(ar);

			long a=0,b=0,c=0;

			a = ar[0]*ar[1]*ar[2]*ar[3]*ar[n-1];
			b = ar[0]*ar[1]*ar[n-1]*ar[n-2]*ar[n-3] ;
			c = ar[n-1]*ar[n-2]*ar[n-3]*ar[n-4]*ar[n-5] ;

			if(a>=b && a>c)
			System.out.println(a);
			else if (b>a && b>c)
				System.out.println(b);
			else
				System.out.println(c);
		}
		}
	}
}
			
