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;
  void _print(long long t) { cout << t; }
  void _print(int t) { cout << t; }
  void _print(string t) { cout << t; }
  void _print(char t) { cout << t; }
  void _print(long double t) { cout << t; }
  void _print(double t) { cout << t; }
  void _print(unsigned long long t) { cout << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cout << "{";
    _print(p.first);
    cout << ",";
    _print(p.second);
    cout << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cout << "[ ";
    for (T i : v) {
      _print(i);
      cout << " ";
    }
    cout << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cout << "[ ";
    for (T i : v) {
      _print(i);
      cout << " ";
    }
    cout << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cout << "[ ";
    for (T i : v) {
      _print(i);
      cout << " ";
    }
    cout << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cout << "[ ";
    for (auto i : v) {
      _print(i);
      cout << " ";
    }
    cout << "]";
  }
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    cout << name << " : " << arg1 << endl;
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    cout.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  long long gcd(long long a, long long b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long expo(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % (int)(1e9 + 7);
      a = (a * a) % (int)(1e9 + 7);
      b = b >> 1;
    }
    return res;
  }
  vector<long long> sieve(int n) {
    int* arr = new int[n + 1]();
    vector<long long> vect;
    for (int i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
      }
    return vect;
  }
  void solve() {
    cout << setprecision(10) << fixed;
    int n;
    cin >> n;
    int a[n];
    long long s = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      s += a[i];
    }
    sort(a, a + n);
    cout << (s - a[n - 1]) / double(n - 1) + a[n - 1] << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++) solve();
  }
  