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 Main{
  public static void main(String[] args)
  {
    Scanner sc=new Scanner(System.in);
    int t=sc.nextInt();
    for(int z=0;z<t;z++)
    {
      int n=sc.nextInt();
      long a[]=new long[n];
      for(int i=0;i<n;i++)
      {
        a[i]=sc.nextLong();
      }
      if(n==5)
      {
        long as=1;
        for(int i=0;i<5;i++)
        as*=a[i];
        System.out.println(as);
        continue;
      }
      Arrays.sort(a);
      if(a[n-1]<0)
      {
        long as=1;
        for(int i=0;i<5;i++)
        as*=a[n-i-1];
        System.out.println(as);
        continue;
      }
      long ans=a[n-1];
      int beg=0,end=n-2;
      for(int i=0;i<2;i++)
      {
        if(a[beg]*a[beg+1]>a[end]*a[end-1])
        {
          ans=ans*a[beg]*a[beg+1];
          beg+=2;
        }
        else
        {
          ans=ans*a[end]*a[end-1];
          end-=2;
        }
      }
      System.out.println(ans);
    }
  }
}

  