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;

public class MaximumProduct {
    public static void  main(String[] args) {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        while (t-- > 0) {
            int n = s.nextInt();
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = s.nextLong();
            }
            Arrays.sort(a);
            int positiveCount=0;
            int negativeCount=0;
            boolean zeroExists=false;
            for(int i=n-1;i>=0;i--)
            {
                if(a[i]>0) positiveCount++;
                else if(a[i]<0) negativeCount++;
                else zeroExists=true;
            }
            long finalResult=1;
            if(positiveCount>=5) {
                finalResult = finalResult * a[n - 1];
                if (a[n - 2] * a[n - 3] > a[0] * a[1]) {
                    finalResult = finalResult * a[n - 3] * a[n - 2];
                    if (a[n - 4] * a[n - 5] > a[0] * a[1]) finalResult = finalResult * a[n - 4] * a[n - 5];
                    else finalResult = finalResult * a[0] * a[1];
                } else {
                    finalResult = finalResult * a[0] * a[1];
                    if (a[2] * a[3] > a[n - 2] * a[n - 3]) finalResult = finalResult * a[2] * a[3];
                    else finalResult = finalResult * a[n - 3] * a[n - 2];
                }

            }
            else if(positiveCount==4)
            {
                if(negativeCount==1 || negativeCount==0)
                {
                    if(zeroExists)finalResult=0;
                    else{
                        for(int i=0;i<5;i++)finalResult=finalResult*a[n-1-i];
                    }
                }
                else{
                    if(negativeCount==2 || negativeCount==3)finalResult=a[n-1]*a[n-2]*a[n-3]*a[0]*a[1];
                    else{
                        finalResult=a[n-1]*a[0]*a[1];
                        if(a[2]*a[3]>a[n-2]*a[n-3])finalResult=finalResult*a[2]*a[3];
                        else finalResult=finalResult*a[n-2]*a[n-3];
                    }


                }

            }
            else if(positiveCount==3){
                if(negativeCount<2)finalResult=0;
                else{
                    if(negativeCount<=3)
                    {
                        finalResult=a[n-1]*a[n-2]*a[n-3]*a[0]*a[1];
                    }
                    else{
                        finalResult=a[n-1]*a[0]*a[1];
                        if(a[2]*a[3]>a[n-2]*a[n-3])finalResult=finalResult*a[2]*a[3];
                        else finalResult=finalResult*a[n-2]*a[n-3];
                    }

                }
            }
            else if(positiveCount==2){
                if(negativeCount<3)finalResult=0;
                else{
                    if(negativeCount==3 && zeroExists)finalResult=0;
                    else finalResult=a[n-1]*a[0]*a[1]*a[2]*a[3];
                }
            }
            else if(positiveCount==1) {
                if (negativeCount < 4) finalResult = 0;
                else {
                    finalResult = a[n - 1] * a[0] * a[1] * a[2] * a[3];
                }
            }
            else{
                for(int i=0;i<5;i++)finalResult=finalResult*a[n-1-i];
            }
            System.out.println((finalResult));
        }
    }
}
