#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;
}

### Prompt

Create a solution in CPP for the following problem:

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;
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;
}

```