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 Solution { 
	static long maxprod;
	public static void prod(List<Integer> li,int i,long prod,int count){
	
		if(i>=li.size()) {
			if(count==5) {
				maxprod=Math.max(maxprod,prod);
			}
			return;
		}
		if(count==5) {
			maxprod=Math.max(maxprod,prod);
			return;
			
		}
		
		
		prod(li,i+1,prod,count);
		prod(li,i+1,prod*li.get(i),count+1);
		
		
	}
  
    

      

    public static void main (String[] args) 
    { 
    	Scanner sc=new Scanner(System.in);
    	int t=sc.nextInt();
    	for(int ii=0;ii<t;ii++) {
    	int n=sc.nextInt();
        int arr[] = new int[n];
        
        for(int i=0;i<n;i++) arr[i]=sc.nextInt();
        Arrays.sort(arr);
         List<Integer> li=new ArrayList<>();
         for(int i=0;i<n;i++) {
        	 if(i<5||i>=n-6) {
        		 li.add(arr[i]);
        	 }
         }
         

    	
         maxprod=Long.MIN_VALUE;
         prod(li,0,1,0);
         System.out.println(maxprod);
         
      
    	}
    } 
} 