Problem: The heat during the last few days has been really intense. Scientists from all over the Berland study how the temperatures and weather change, and they claim that this summer is abnormally hot. But any scientific claim sounds a lot more reasonable if there are some numbers involved, so they have decided to actually calculate some value which would represent how high the temperatures are.

Mathematicians of Berland State University came up with a special heat intensity value. This value is calculated as follows:

Suppose we want to analyze the segment of n consecutive days. We have measured the temperatures during these n days; the temperature during i-th day equals a_i.

We denote the average temperature of a segment of some consecutive days as the arithmetic mean of the temperature measures during this segment of days. So, if we want to analyze the average temperature from day x to day y, we calculate it as \frac{∑ _{i = x}^{y} a_i}{y - x + 1} (note that division is performed without any rounding). The heat intensity value is the maximum of average temperatures over all segments of not less than k consecutive days. For example, if analyzing the measures [3, 4, 1, 2] and k = 3, we are interested in segments [3, 4, 1], [4, 1, 2] and [3, 4, 1, 2] (we want to find the maximum value of average temperature over these segments).

You have been hired by Berland State University to write a program that would compute the heat intensity value of a given period of days. Are you up to this task?

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 5000) — the number of days in the given period, and the minimum number of days in a segment we consider when calculating heat intensity value, respectively.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 5000) — the temperature measures during given n days.

Output

Print one real number — the heat intensity value, i. e., the maximum of average temperatures over all segments of not less than k consecutive days.

Your answer will be considered correct if the following condition holds: |res - res_0| < 10^{-6}, where res is your answer, and res_0 is the answer given by the jury's solution.

Example

Input

4 3
3 4 1 2


Output

2.666666666666667

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void c_p_c() {}
class DisjointSet {
  unordered_map<int, int> parent;

 public:
  void makeSet(int N) {
    for (int i = 0; i < N; i++) parent[i] = i;
  }
  int Find(int v) {
    if (v == parent[v]) return v;
    return parent[v] = Find(parent[v]);
  }
  void Union(int a, int b) {
    int x = Find(a);
    int y = Find(b);
    parent[x] = y;
  }
};
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void PrimeFactors(long long n) {
  while (n % 2 == 0) {
    cout << 2 << " ";
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      cout << i << " ";
      n = n / i;
    }
  }
  if (n > 2) cout << n << " ";
}
template <typename T>
void parray(T a[], T n) {
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
vector<long long> uniqueelement(vector<long long> &a) {
  auto ip = unique((a).begin(), (a).end());
  a.resize(distance(a.begin(), ip));
  return a;
}
bool IsPrime(long long x) {
  if (x == 1) return false;
  for (int i = 2; 1ll * i * i <= x; i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
void leftrotate(string &s, int d) {
  reverse(s.begin(), s.begin() + d);
  reverse(s.begin() + d, s.end());
  reverse(s.begin(), s.end());
}
void rightrotate(string &s, int d) { leftrotate(s, s.length() - d); }
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
int XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
vector<long long> sieve() {
  vector<long long> primes;
  bool a[1000000] = {0};
  int mprimes = 1000000;
  a[0] = a[1] = true;
  for (int i = 2; i * i < mprimes; ++i) {
    if (a[i] == false) {
      for (int j = i * i; j < mprimes; j = j + i) {
        a[j] = true;
      }
    }
  }
  for (int i = 2; i < mprimes; ++i) {
    if (!a[i]) primes.push_back(i);
    ;
  }
  return primes;
}
void Ynot() {
  long long i, n, m, k, c = 0, sum = 0;
  bool flag = false;
  cin >> n >> k;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  ;
  for (int i = 1; i < n; ++i) {
    a[i] = a[i] + a[i - 1];
  }
  double ans = 0;
  while (k <= n) {
    int i = 0, j = k - 1;
    for (int i = 0; j < n; ++i) {
      double ans1 = 0;
      if (i == 0) {
        ans1 = a[j];
      } else {
        ans1 = a[j] - a[i - 1];
      }
      ans1 = ans1 / k;
      ans = max(ans1, ans);
      j++;
    }
    k++;
  }
  cout << fixed << setprecision(6) << ans << "\n";
}
int main() {
  c_p_c();
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) {
    Ynot();
  }
  return 0;
}