#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first != b.first.first) return a.first.first > b.first.first;
  if (a.first.second != b.first.second) return a.first.second < b.first.second;
  return a.second < b.second;
}
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; }
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 << "]";
}
bool compare(const pair<int, int>& i, const pair<int, int>& j) {
  return i.second < j.second;
}
void solve() {
  int n;
  cin >> n;
  vector<long long> v(n);
  bool all_neg = true;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] > 0) {
      all_neg = false;
    }
  }
  long long sum = 0, ans = 0;
  sort(v.begin(), v.end());
  ;
  ;
  sort(v.begin(), v.end());
  for (int i = 0; i < n - 1; i++) {
    sum += v[i];
  };
  ;
  cout << fixed << setprecision(9) << v[n - 1] + (double)sum / (n - 1) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first != b.first.first) return a.first.first > b.first.first;
  if (a.first.second != b.first.second) return a.first.second < b.first.second;
  return a.second < b.second;
}
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; }
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 << "]";
}
bool compare(const pair<int, int>& i, const pair<int, int>& j) {
  return i.second < j.second;
}
void solve() {
  int n;
  cin >> n;
  vector<long long> v(n);
  bool all_neg = true;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] > 0) {
      all_neg = false;
    }
  }
  long long sum = 0, ans = 0;
  sort(v.begin(), v.end());
  ;
  ;
  sort(v.begin(), v.end());
  for (int i = 0; i < n - 1; i++) {
    sum += v[i];
  };
  ;
  cout << fixed << setprecision(9) << v[n - 1] + (double)sum / (n - 1) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```