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.ArrayList;
  import java.util.Collection;
  import java.util.Collections;
  import java.util.List;
  import java.util.Scanner;
  import java.util.StringTokenizer;
  import java.util.*;
  
  public class A {
      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();
          }
  
          ArrayList<Integer> readIntArrayList(int n) {
              ArrayList<Integer> a = new ArrayList<>();
              for (int i = 0; i < n; i++)
                  a.add(nextInt());
              return a;
          }
  
          ArrayList<Long> readLongArrayList(int n) {
              ArrayList<Long> a = new ArrayList<>();
              for (int i = 0; i < n; i++)
                  a.add(nextLong());
              return a;
          }
  
          int[] readIntArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextInt();
              return a;
          }
  
          long[] readLongArray(int n) {
              long[] a = new long[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextInt();
              return a;
          }
  
          char[] readCharArray(int n) {
              char[] a = new char[n];
              for (int i = 0; i < n; i++)
                  a[i] = next().charAt(0);
              return a;
          }
  
          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;
          }
      }
  
      static Boolean[] sieveOfEratosthenes(int n, Boolean[] prime) {
          for (int i = 0; i <= n; i++)
              prime[i] = true;
  
          for (int p = 2; p * p <= n; p++) {
              // If prime[p] is not changed, then it is a
              // prime
              if (prime[p] == true) {
                  // Update all multiples of p
                  for (int i = p * p; i <= n; i += p)
                      prime[i] = false;
              }
          }
          return prime;
  
          // Print all prime numbers
      }
  
      static int gcd(int a, int b) {
          return b == 0 ? a : gcd(b, a % b);
      }
  
      static int isPrime(int n) {
          for (int i = 2; i * i <= n; i++) {
              if (n % i == 0) {
                  return i;
              }
          }
          return -1;
      }
  
      static void p(String s) {
          System.out.print(s);
      }
  
      static void pln(String s) {
          System.out.println(s);
      }
  
      static String ts(int x) {
          return Integer.toString(x);
      }
  
      static boolean searchX(int x, int n, int a, int b) {
          if (x == n) {
              return true;
          } else if (x > n) {
              return false;
          } else {
              return (a != 1) ? searchX(x * a, n, a, b) || searchX(x + b, n, a, b) : searchX(x + b, n, a, b);
          }
      }
  
      static boolean check(int[][] grid, int i, int j, int h, int w) {
          int dx[] = { 0, 0, 1, -1, 1, 1, -1, -1 };
          int dy[] = { 1, -1, 0, 0, 1, -1, 1, -1 };
  
          for (int d = 0; d < 8; d++) {
              if (i + dx[d] < 0 || i + dx[d] >= h || j + dy[d] < 0 || j + dy[d] >= w)
                  continue;
              else {
                  if (grid[i + dx[d]][j + dy[d]] == 1)
                      return false;
              }
          }
  
          return true;
      }
  
      static int iterate(int[][] grid, int r, int c, int h, int w) {
          int count = 0;
  
          for (int i = r; i < h; i++) {
              for (int j = c; j < w; j++) {
                  if (check(grid, i, j, h, w)) {
                      grid[i][j] = 1;
                      count += 1;
                  }
              }
          }
          // for (int i = 0; i < h; i++) {
          // for (int j = 0; j < w; j++) {
          // System.out.print(grid[i][j] + " ");
          // }
          // System.out.println();
          // }
          return count;
  
      }
  
      public static void main(String[] args) {
  
          FastReader fr = new FastReader();
          int t = fr.nextInt();
          while (t-- > 0) {
              int n = fr.nextInt();
              long arr[] = fr.readLongArray(n);
              Arrays.sort(arr);
              long sum = 0;
              for (long i : arr)
                  sum += i;
              double p1 = (double) arr[n - 1];
              double p2 = (double) (sum - arr[n - 1]) / (double) (n - 1);
  
              System.out.println(p1 + p2);
  
          }
      }
  }