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;

/**
 * @author rishav
 *
 */
/**
 * @author rishav
 *
 */
public class B670 {

	

	public static void main(String[]args) {
		Scanner sc=new Scanner(System.in);
	        int t=sc.nextInt();
	        while(t>0) {
			int  n=sc.nextInt();
		    long arr[]=new long[n];
		    for(int i=0;i<n;i++) {
		    	arr[i]=sc.nextLong();
		    }
			System.out.println(solve(arr,n));
			t--;
	        }
}
	static long solve(long[]a,int n) {
		long res=1;
		if(n==5) {
			res=a[0]*a[1]*a[2]*a[3]*a[4];
		    return res;
		}
		else {
			
			Arrays.sort(a);
			if(a[0]>=0) {
				for(int i=n-1;i>=n-5;i--) {
					res*=a[i];
				}
				return res;
			}
			else if(a[n-1]<=0) {
				for(int i=n-1;i>=n-5;i--) {
					res*=a[i];
					
				}
				return res;
			}
			
				
				
			else {
				
				long op1=Math.max(a[0]*a[1]*a[2]*a[3]*a[n-1], a[0]*a[1]*a[n-1]*a[n-2]*a[n-3]);
				 res=Math.max(op1, a[n-1]*a[n-2]*a[n-3]*a[n-4]*a[n-5]);
				
				return res;
			
		}
		}
	}
}


