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.*;
 
public class Solution {
    static long mod = 1000000007;
    public static void main(String[] args) {
 
        Scanner in = new Scanner(System.in);
        int t=in.nextInt();
        while(t-->0){
            int n=in.nextInt();
            Long[] a=new Long[n];
            long maxV=Integer.MIN_VALUE;
            for(int i=0;i<n;i++){
                a[i]=in.nextLong();
                maxV=Math.max(maxV,a[i]);
            }
            Arrays.sort(a,(o1,o2)-> Long.compare(Math.abs(o2), Math.abs(o1)));
            long ans=1L*a[n-1]*a[n-2]*a[n-3]*a[n-4]*a[n-5];
            if(maxV<0){
                System.out.println(ans);
                continue;
            }
 
            long max=1L*a[0]*a[1]*a[2]*a[3]*a[4];
            for(int i=5;i<n;i++){
                for(int j=0;j<5;j++){
                    long tempMax= a[i];
                    for(int k=0;k<5;k++){
                        if(k!=j)
                            tempMax*=a[k];
                    }
                    max=Math.max(tempMax,max);
                }
            }
            // if(max==-384){
            //     System.out.println(Arrays.toString(a));
            // }
            System.out.println(max);

        }
 
    }
 
}
