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;
  const int MOD = 1e9 + 7;
  long long int binPow(long long int a, long long int n) {
    long long int res = 1;
    while (n) {
      if (n & 1) res = (1LL * res * a) % MOD;
      a = (1LL * a * a) % MOD;
      n >>= 1;
    }
    return res;
  }
  long long int gcd(long long int a, long long int b) {
    while (b) {
      a %= b;
      swap(a, b);
    }
    return a;
  }
  long long int lcm(long long int a, long long int b) {
    return (a / gcd(a, b)) * b;
  }
  bool prime(long long int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    for (long long int i = 5; i * i <= n; i = i + 6) {
      if (n % i == 0 || n % (i + 2) == 0) {
        return false;
      }
    }
    return true;
  }
  long long int ncr(long long int n, long long int k,
                    const vector<long long int>& fact,
                    const vector<long long int>& inv) {
    if (k > n) return 0;
    long long int m = (1LL * fact[n] * inv[k]) % MOD;
    m = (1LL * m * inv[n - k]) % MOD;
    return m;
  }
  void solve() {
    long long int n, s = 0;
    cin >> n;
    vector<long long int> v(n);
    for (long long int i = 0; i < n; i++) cin >> v[i];
    sort(v.begin(), v.end());
    for (long long int i = 0; i < n - 1; i++) {
      s += v[i];
    }
    double avg = (double)s / (n - 1);
    avg += v[n - 1];
    cout << fixed << setprecision(9) << avg << endl;
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  