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.math.BigInteger;
  import java.util.*;
  import java.io.*;
  
  public class A_Ezzat_and_Two_Subsequences {
      public static void main (String[] args) {
          FastReader scan = new FastReader();
          int t = scan.nextInt();
          while (t-- > 0) {
              int n = scan.nextInt();
              int[] arr = new int[n];
              inputArray(arr, n, scan);
              int max = arr[0], maxi = 0;
              for (int i=0; i<n; i++) {
                  if (max < arr[i]) {
                      max = arr[i];
                      maxi = i;
                  }
              }
              double ans = max, sum = 0;
              for (int i=0; i<n; i++) {
                  if (i != maxi) {
                      sum += arr[i];
                  }
              }
              ans += sum/(n-1);
              System.out.printf("%.9f\n", ans);
          }
      }
      public static void inputArray (int[] arr, int n, FastReader scan) {
          for (int i=0; i<n; i++) {
              arr[i] = scan.nextInt();
          }
      }
      public static void printArray (int[] arr) {
          for (int i=0; i<arr.length; i++) {
              System.out.print(arr[i] + " ");
          }
          System.out.println();
      }
      public static boolean isPrime (int n) {
          if (n <= 1)
              return false;
          if (n <= 3)
              return true;
          if (n % 2 == 0 || n % 3 == 0)
              return false;
          for (int i = 5; i * i <= n; i = i + 6)
              if (n % i == 0 || n % (i + 2) == 0)
                  return false;
          return true;
      }
      public static int gcd (int a, int b) {
          BigInteger a1 = BigInteger.valueOf(a);
          BigInteger b1 = BigInteger.valueOf(b);
          return a1.gcd(b1).intValue();
      }
      public static int numberOfDigitsInNumber (int n) {
          return (int)Math.floor(Math.log10(n)) + 1;
      }
      public static boolean isPowerOfTwo (int x) {
          return x!=0 && ((x&(x-1)) == 0);   
      }
      public static String toBinaryString (int n) {
          return Integer.toString(n, 2);
          // for other bases change 2 to other bases
      }
      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;
          }
      }
  }
  
  // For IO using Files
  // public static void main throws IOException
  // BufferedReader br = new BufferedReader(new FileReader("input.txt"));
  // PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
  // String st[]=br.readLine().split(" ");
  // int n=Integer.parseInt(st[0]);
  // int m=Integer.parseInt(st[1]);
  // at the end
  // br.close(); pw.close();
  // for printing pw.println(), pw.print(), etc...
  // for arrays : int arr[] = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();