description: Ezzat has an array of n integers (maybe negative). He wants to split it into two non-empty subsequences a and b, such that every element from the array belongs to exactly one subsequence, and the value of f(a) + f(b) is the maximum possible value, where f(x) is the average of the subsequence x. 
  
  A sequence x is a subsequence of a sequence y if x can be obtained from y by deletion of several (possibly, zero or all) elements.
  
  The average of a subsequence is the sum of the numbers of this subsequence divided by the size of the subsequence.
  
  For example, the average of [1,5,6] is (1+5+6)/3 = 12/3 = 4, so f([1,5,6]) = 4.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^3)— the number of test cases. Each test case consists of two lines.
  
  The first line contains a single integer n (2 ≤ n ≤ 10^5).
  
  The second line contains n integers a_1, a_2, …, a_n (-10^9 ≤ a_i ≤ 10^9).
  
  It is guaranteed that the sum of n over all test cases does not exceed 3⋅10^5.
  
  Output
  
  For each test case, print a single value — the maximum value that Ezzat can achieve.
  
  Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
  
  Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.
  
  Example
  
  Input
  
  
  4
  3
  3 1 2
  3
  -7 -6 -6
  3
  2 2 2
  4
  17 3 5 -3
  
  
  Output
  
  
  4.500000000
  -12.500000000
  4.000000000
  18.666666667
  
  Note
  
  In the first test case, the array is [3, 1, 2]. These are all the possible ways to split this array: 
  
    * a = [3], b = [1,2], so the value of f(a) + f(b) = 3 + 1.5 = 4.5. 
    * a = [3,1], b = [2], so the value of f(a) + f(b) = 2 + 2 = 4. 
    * a = [3,2], b = [1], so the value of f(a) + f(b) = 2.5 + 1 = 3.5. 
  
  Therefore, the maximum possible value 4.5.
  
  In the second test case, the array is [-7, -6, -6]. These are all the possible ways to split this array: 
  
    * a = [-7], b = [-6,-6], so the value of f(a) + f(b) = (-7) + (-6) = -13. 
    * a = [-7,-6], b = [-6], so the value of f(a) + f(b) = (-6.5) + (-6) = -12.5. 
  
  Therefore, the maximum possible value -12.5.
solution: import java.io.*;
  import java.util.*;
  public class Main {
      public static void main(String args[])
      {
          FastReader input=new FastReader();
          PrintWriter out=new PrintWriter(System.out);
          int T=input.nextInt();
          while(T-->0)
          {
              int n=input.nextInt();
              int a[]=new int[n];
              long pre[]=new long[n];
              long sum=0;
              for(int i=0;i<n;i++)
              {
                  a[i]=input.nextInt();
              }
              mergeSort(a,0,n-1);
              for(int i=0;i<n;i++)
              {
                  sum+=a[i];
                  pre[i]=sum;
              }
              double max=-1000000000000000d;
              for(int i=0;i<n-1;i++)
              {
                  double s1=pre[i];
                  double s2=pre[n-1]-pre[i];
                  double av1=s1/(i+1);
                  double av2=s2/(n-i-1);
                  double av=av1+av2;
                  max=Math.max(max,av);
              }
              out.println(max);
          }
          out.close();
      }
      public static void mergeSort(int a[],int p,int r)
      {
          if(p<r)
          {
              int q=(p+r)/2;
              mergeSort(a,p,q);
              mergeSort(a,q+1,r);
              merge(a,p,q,r);
          }
      }
      public static void merge(int a[],int p,int q,int r)
      {
          int n1=q-p+2;
          int L[]=new int[n1];
          int n2=r-q+1;
          int R[]=new int[n2];
          for(int i=p;i<=q;i++)
          {
              L[i-p]=a[i];
          }
          L[n1-1]=Integer.MIN_VALUE;
          for(int i=q+1;i<=r;i++)
          {
              R[i-q-1]=a[i];
          }
          R[n2-1]=Integer.MIN_VALUE;
          int x=0,y=0;
          for(int i=p;i<=r;i++)
          {
              if(L[x]>=R[y])
              {
                  a[i]=L[x];
                  x++;
              }
              else
              {
                  a[i]=R[y];
                  y++;
              }
          }
      }
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer st;
          public FastReader()
          {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
          String next()
          {
              while (st == null || !st.hasMoreElements())
              {
                  try
                  {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e)
                  {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
          int nextInt()
          {
              return Integer.parseInt(next());
          }
          long nextLong()
          {
              return Long.parseLong(next());
          }
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
          String nextLine()
          {
              String str="";
              try
              {
                  str=br.readLine();
              }
              catch (IOException e)
              {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }