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: #include <bits/stdc++.h>
  using namespace std;
  int power(int a, int b) {
    if (b == 0) {
      return 1;
    } else if (b == 1) {
      return a;
    } else if (b % 2 == 0) {
      int x = power(a, b / 2);
      return x * x;
    } else {
      int x = power(a, (b - 1) / 2);
      return x * x * a;
    }
  }
  bool prime(int n) {
    int flag = 0, i;
    if (n == 1) {
      return false;
    } else if (n == 2) {
      return true;
    } else {
      for (i = 2; i <= (int)sqrt(n); i++) {
        if (n % i == 0) {
          flag = 1;
          break;
        }
      }
      if (flag == 1) {
        return false;
      } else {
        return true;
      }
    }
  }
  bool palindrome(string a) {
    string b;
    b = a;
    reverse(b.begin(), b.end());
    if (a == b) {
      return true;
    } else {
      return false;
    }
  }
  bool perfect_square(int num) {
    double a = sqrt(num);
    int b = (int)a;
    if (b * b == num) {
      return true;
    } else {
      return false;
    }
  }
  int binary_search(int* arr, int n, int element) {
    sort(arr, arr + n);
    int low = 0, high = n - 1;
    int mid;
    while (high - low > 1) {
      mid = (low + high) / 2;
      if (arr[mid] < element) {
        low = mid + 1;
      } else {
        high = mid;
      }
    }
    if (arr[low] == element) {
      return low;
    } else if (arr[high] == element) {
      return high;
    } else {
      return -1;
    }
  }
  void solve() { cout << "" << endl; }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      long long a[n];
      long long sum = 0;
      long long maxx = LLONG_MIN;
      for (long long i = 0; i < n; i++) {
        cin >> a[i];
        sum += a[i];
        maxx = max(maxx, a[i]);
      }
      sum = sum - maxx;
      double ans = (double)(sum / ((double)(n - 1))) + (double)maxx;
      printf("%.9lf\n", ans);
    }
    return 0;
  }
  