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) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << 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) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <typename T, typename T1>
  T amax(T &a, T1 b) {
    if (b > a) a = b;
    return a;
  }
  template <typename T, typename T1>
  T amin(T &a, T1 b) {
    if (b < a) a = b;
    return a;
  }
  void FASTIO() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  template <typename P>
  P pow(P x, P n) {
    if (n == 0) return 1;
    if (n % 2 == 0) {
      P y = pow(x, n / 2);
      return y * y;
    } else
      return x * pow(x, n - 1);
  }
  long long modpow(long long x, long long n) {
    if (n == 0)
      return 1;
    else if (n % 2 == 0) {
      long long y = modpow(x, n / 2);
      return (y * y) % 1000000007;
    } else
      return ((x % 1000000007) * modpow(x, n - 1) % 1000000007);
  }
  int gcd(int a, int b) { return (b == 0) ? a : gcd(b, a % b); }
  int countBits(int number) { return (int)log2(number) + 1; }
  int numlen(int a) {
    int res = 0;
    while (a > 0) {
      res++;
      a /= 10;
    }
    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() {
    long long n;
    cin >> n;
    vector<long long int> arr(n);
    for (int i = 0; i < n; i++) cin >> arr[i];
    double ans;
    long long mx = LLONG_MIN;
    long long indx = -1;
    for (int i = 0; i < n; i++) {
      if (mx < arr[i]) {
        mx = arr[i];
        indx = i;
      }
    }
    long long other = 0;
    for (int i = 0; i < n; i++) {
      if (i != indx) other += arr[i];
    }
    ans = ((double)other / (n - 1));
    ans += mx;
    cout << fixed << setprecision(10) << ans;
    cout << endl;
  }
  int main() {
    FASTIO();
    int t;
    t = 1;
    cin >> t;
    for (int test = 1; test <= t; test++) {
      solve();
    }
    return 0;
  }
  