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;
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
long long power(long long x, long long n) {
  if (x == 0) return 0;
  if (n == 0) return 1;
  if (n == 1) return x;
  long long res = power(x, n / 2);
  res = (res * res) % 1000000007;
  if (n % 2) return (res * x) % 1000000007;
  return res;
}
long long nCr(long long n, long long r) {
  if (r == 0) return 1;
  vector<long long> f(n + 1);
  f[0] = 1;
  for (long long i = 1; i <= n; i++) f[i] = (f[i - 1] * i) % 1000000007;
  return ((f[n] * power(f[r], 1000000007 - 2)) % 1000000007 *
          power(f[n - r], 1000000007 - 2)) %
         1000000007;
}
const long long MAX = 1000001;
long long factor[MAX] = {0};
void gen_Prime_Fac() {
  factor[1] = 1;
  for (long long i = 2; i < MAX; i++) factor[i] = i;
  for (long long i = 4; i < MAX; i += 2) factor[i] = 2;
  for (long long i = 3; i * i < MAX; i++) {
    if (factor[i] == i) {
      for (long long j = i * i; j < MAX; j += i) {
        if (factor[j] == j) factor[j] = i;
      }
    }
  }
}
long long no_of_fac(long long n) {
  if (n == 1) return 1;
  long long ans = 1;
  long long dup = factor[n];
  long long c = 1;
  long long j = n / factor[n];
  while (j != 1) {
    if (factor[j] == dup)
      c += 1;
    else {
      dup = factor[j];
      ans = ans * (c + 1);
      c = 1;
    }
    j /= factor[j];
  }
  ans *= (c + 1);
  return ans;
}
bool is_pow_2(long long n) {
  long long i = 1;
  while (i < n) {
    i *= 2;
  }
  return i == n;
}
int32_t main() {
  long long n, k;
  cin >> n >> k;
  vector<long double> pref(n);
  long double sum = 0;
  for (long long i = 0; i < n; i++) {
    long double x;
    cin >> x;
    sum += x;
    pref[i] = sum;
  }
  long double ans = 0;
  for (long long i = k; i <= n; i++) {
    long double sum = pref[i - 1];
    for (long long j = i; j < n; j++) {
      sum = max(sum, (pref[j] - pref[j - i]));
    }
    ans = max(ans, sum / i);
  }
  cout << fixed << setprecision(10) << ans;
}