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.
//'main' method must be in a class 'Rextester'.
//openjdk version '11.0.5' 

import java.util.*;
import java.lang.*;

public class Rextester
{  
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0)
        {
            int n=sc.nextInt();
            int a[]=new int[n];
            int pos=0,neg=0,zero=0;
            for(int i=0;i<n;i++)
            {
                a[i]=sc.nextInt();
                if(a[i]>0)
                    pos++;
                else if(a[i]<0)
                    neg++;
                    else
                    zero++;
            }
            int temp=0;
            long prod=0;
            Arrays.sort(a);
            long prod1=0,prod2=0;
            /*for(int i=0;i<n;i++)
                System.out.print(a[i]+" ");
            System.out.println();*/
            long z=a[n-1];
            long b=a[0]*a[1];
            long c=a[n-2]*a[n-3];
            long d=b*a[2]*a[3];
            long e=c*a[n-4]*a[n-5];
            long mx=(Math.max(z*b*c,Math.max(z*d,e*z)));
            System.out.println(mx);
 
            /*if(pos>=2)
            {
                prod=a[0]*a[1]*a[n-1]*a[n-2]*a[n-3];
                prod1=a[0]*a[1]*a[2]*a[3]*a[4]*a[n-1];
                prod2=a[n-1]*a[n-2]*a[n-3]*a[n-4]*a[n-5];
                prod1=Math.max(prod1,prod2);
                prod=Math.max(prod,prod1);
            }
            else if(pos>=1&&neg>=4)
            {
                prod=a[0]*a[1]*a[2]*a[3]*a[n-1];
            }
            else if(pos==0&&zero!=0)
                prod=0;
            else if(pos==0&&zero==0)
                prod=a[n-1]*a[n-2]*a[n-3]*a[n-4]*a[n-5];
            System.out.println(prod);*/
                
                
                
        }
    }
}