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.util.*;
  import java.lang.*;
  import java.io.*;
  
  //author: ankit_bit
  
  public class Main {
  
      static FastReader fr = new FastReader();
      static StringBuilder sb = new StringBuilder();
  
      public static void main(String[] args) throws java.lang.Exception {
  
          int testcases = fr.nextInt();
          while (testcases-- > 0) {
  
              solve();
          }
          System.out.println(sb);
  
      }
  
      public static void solve() {
          int n= fr.nextInt();
          double a[] = new double[n];
          for (int i = 0; i < a.length; i++) {
              a[i]= fr.nextDouble();
          }
  
          Arrays.sort(a);
          double x = a[n-1];
  
          double y=0;
          for (int i = 0; i < n-1; i++) {
              y+=a[i];
          }
  
          y=y/(n-1);
  
          sb.append(x+y);
                sb.append("\n");
      }
  
  
  
      static boolean isPrime(int x) {
          for (int d = 2; d * d <= x; d++) {
              if (x % d == 0)
                  return false;
          }
          return true;
      }
  
      static int lowerBound(int a[], int low, int high, int x) {
          int pos = -1;
          while (low <= high) {
              int mid = (low + high) / 2;
              if (a[mid] == x) {
                  pos = mid;
                  high = mid - 1;
              } else if (a[mid] < x) low = mid + 1;
              else high = mid - 1;
          }
          if (pos != -1) return pos;
          return high + 1;
      }
  
      static int upperBound(int a[], int low, int high, int x) {
          int pos = -1;
          while (low <= high) {
              int mid = (low + high) / 2;
              if (a[mid] == x) {
                  pos = mid;
                  low = mid + 1;
              } else if (a[mid] < x) low = mid + 1;
              else high = mid - 1;
          }
          if (pos != -1) return pos;
          return high + 1;
      }
  
      static long mult(long a, long b, long mod) {
          long res = (a * b) % mod;
          if (res < 0) {
              res += mod;
          }
          return res;
      }
  
      static long add(long a, long b, long mod) {
          long res = (a + b) % mod;
          if (res < 0) {
              res += mod;
          }
          return res;
      }
  
      static long gcd(long a, long b) {
          if (b == 0) return a;
          return gcd(b, a % b);
      }
  
      static long lcm(long a, long b) {
          return (a / gcd(a, b)) * b;
      }
  
      static long fact[] = new long[100001], M = 1000000007;
  
      static long ncr(int n, int r) {
          if (r > n) return 0;
          return fact[n] * power(fact[r], M - 2) % M * power(fact[n - r], M - 2) % M;
      }
  
      static long power(long a, long b) {
          long res = 1;
          while (b != 0) {
              if (b % 2 == 1) res = res * a % M;
              b >>= 1;
              a = a * a % M;
          }
          return res;
      }
  
  
      static void sort(int[] a) {
          ArrayList<Integer> l = new ArrayList<>();
          for (int i = 0; i < a.length; i++) {
              l.add(a[i]);
          }
          Collections.sort(l);
          for (int i = 0; i < l.size(); i++)
              a[i] = l.get(i);
      }
  
  }
  
  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;
      }
  }
  