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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  import java.lang.*;
  import java.math.*;
  
  public class ArrayQue {
  
      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;
  
          }
  
      }
  
      public static void main(String[] args) {
          FastReader  sc=new FastReader();
          int t=sc.nextInt();
  
          while(t-->0) {
              int n=sc.nextInt();
  
  ArrayList<Long> al=new ArrayList<>();
  
  for(int i=0; i<n; i++) {
  	long x=sc.nextLong();
  	al.add(x);
  }
  boolean flag=false;
  long sl=0,sr=0;
  long max= Collections.max(al);
  long min=Collections.min(al);
  for(int i=0; i<n; i++){
  
          if( flag==false && al.get(i)==max) {
          flag=true;
          }
          else{
          sl+=al.get(i);
  
              }
  }
  double it=(double) max;
  double ti=((double) sl/(double) (long)(n-1));
  double ans=it+ti;
  System.out.println(ans);
  
          } }}