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 MyTemplate {
  
      void run() throws IOException {
          double t = nextDouble();
          while(t-->0){
              double n = nextDouble(),sm = 0;
              double mx = Long.MIN_VALUE;
              for (int i = 0; i < n; i++) {
                  double x = nextDouble();
                  mx = Math.max(mx,x);
                  sm = sm+x;
  
              }
              System.out.println((sm-mx)/(n-1)+mx);
          }
      }
  
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      StringTokenizer in = new StringTokenizer("");
  
  
      // pairs
      class pil implements Comparable<pil> {
          int first;
          long second;
  
          pil(int f, long s) {
              this.first = f;
              this.second = s;
          }
  
          public int compareTo(pil o) {
              if (first != o.first) return Integer.compare(first, o.first);
              return Long.compare(second, o.second);
          }
      }
  
      class pii implements Comparable<pii> {
          int fi;
          int se;
  
          pii(int s, int f) {
              se = f;
              fi = s;
          }
  
          public int compareTo(pii o) {
              if (fi != o.fi) return Integer.compare(fi, o.fi);
              return Integer.compare(se, o.se);
          }
      }
  
      class pll implements Comparable<pll> {
          long first;
          long second;
  
          pll(long f, long s) {
              this.first = f;
              this.second = s;
          }
  
          public int compareTo(pll o) {
              if (first != o.first) return Long.compare(first, o.first);
              return Long.compare(second, o.second);
          }
      }
  
      class pli implements Comparable<pli> {
          long first;
          int second;
  
          pli(long f, int s) {
              this.first = f;
              this.second = s;
          }
  
          public int compareTo(pli o) {
              if (first != o.first) return Long.compare(first, o.first);
              return Integer.compare(second, o.second);
          }
      }
  
      boolean hasNext() throws IOException {
          if (in.hasMoreTokens()) return true;
          String s;
          while ((s = br.readLine()) != null) {
              in = new StringTokenizer(s);
              if (in.hasMoreTokens()) return true;
          }
          return false;
      }
  
      String nextToken() throws IOException {
          while (!in.hasMoreTokens()) {
              in = new StringTokenizer(br.readLine());
          }
          return in.nextToken();
      }
  
      int nextInt() throws IOException {
          return Integer.parseInt(nextToken());
      }
  
      double nextDouble() throws IOException {
          return Double.parseDouble(nextToken());
      }
  
      long nextLong() throws IOException {
          return Long.parseLong(nextToken());
      }
  
      public static void main(String[] args) throws IOException {
          MyTemplate m = new MyTemplate();
          m.run();
          m.out.close();
      }
      ArrayList<Integer> bfs(ArrayList<HashSet<Integer>> adjacencyList,int V,int source){
          boolean[] discovered=new boolean[V];
          Queue<Integer> discoveredNodes = new LinkedList<>();
          ArrayList<Integer> lvlOrder = new ArrayList<>();
  
          //Additional info
          int[] parent = new int[V]; Arrays.fill(parent,-1);
          int[] distance = new int[V]; Arrays.fill(distance,0);
  
  
          //discovered Source
          discoveredNodes.add(source);
          discovered[source] = true;
          //Additional info source
          parent[source] = -1;
          distance[source] = 0;
  
          while(!discoveredNodes.isEmpty()){
              int parentNode = discoveredNodes.poll();
              lvlOrder.add(parentNode);
              for(int childNode: adjacencyList.get(parentNode)){
                  if(!discovered[childNode]){
                      //discovered nbrs that were not already discovered
                      discoveredNodes.add(childNode);
                      discovered[childNode] = true;
  
                      //filling additional info
                      parent[childNode] = parentNode;
                      distance[childNode] = distance[parentNode]+1;
                  }
  
              }
  
          }
          return lvlOrder;
      }
  }