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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class MaximumProduct
{

	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		
		int t = sc.nextInt();
		while(t-->0)
		{
				
			int n = sc.nextInt();
			ArrayList<Long> num = new ArrayList<Long>();
			ArrayList<Long> pos = new ArrayList<Long>();
			ArrayList<Long> neg = new ArrayList<Long>();
		
			for(int i=0;i<n;i++)
			{
				long x = sc.nextLong();
				num.add(x);
				if(x>=0)
					pos.add(x);
				else
					neg.add(x);
			}
		
			Collections.sort(num,Collections.reverseOrder());
			Collections.sort(pos,Collections.reverseOrder());
			Collections.sort(neg);
		
			long p1 = Long.MIN_VALUE;
			long p2 = Long.MIN_VALUE;
			long p3 = Long.MIN_VALUE;
		
			long t1 = 1;
			for(int i=0;i<5;i++)
				t1*=num.get(i);
			p1 = t1;
		
			if(neg.size()>=2 && pos.size()>=3)
			{
				long t2 = 1;
				for(int i=0;i<2;i++)
					t2*=neg.get(i);
				for(int i=0;i<3;i++)
					t2*=pos.get(i);
				p2=t2;
			}
		
			if(neg.size()>=4 && pos.size()>=1)
			{
				long t3 = 1;
				for(int i=0;i<4;i++)
					t3*=neg.get(i);
				for(int i=0;i<1;i++)
					t3*=pos.get(i);
				p3 = t3;
			}
		
			System.out.println(Math.max(p1, Math.max(p2, p3)));
		
		}
		
	}
	
}
