Problem:
Score : 400 points

Problem Statement

We have N dice arranged in a line from left to right. The i-th die from the left shows p_i numbers from 1 to p_i with equal probability when thrown.

We will choose K adjacent dice, throw each of them independently, and compute the sum of the numbers shown. Find the maximum possible value of the expected value of this sum.

Constraints

1 ≤ K ≤ N ≤ 200000

1 ≤ p_i ≤ 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
p_1 ... p_N

Output

Print the maximum possible value of the expected value of the sum of the numbers shown.

Your output will be considered correct when its absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

5 3
1 2 2 4 5

Sample Output 1

7.000000000000

When we throw the third, fourth, and fifth dice from the left, the expected value of the sum of the numbers shown is 7. This is the maximum value we can achieve.

Sample Input 2

4 1
6 6 6 6

Sample Output 2

3.500000000000

Regardless of which die we choose, the expected value of the number shown is 3.5.

Sample Input 3

10 4
17 13 13 12 15 20 10 13 17 11

Sample Output 3

32.000000000000

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;


#define FOR(i, start, end) for(uint64_t i=start; i<end; i++)
#define REP(i, n) FOR(i, 0, n)

// 最大公約数gcd
// 最小公倍数lcm=m*n/gcd
uint64_t gcd(uint64_t m, uint64_t n) {
  uint64_t temp;
  while (m % n != 0){
    temp = n;
    n = m % n;
    m = temp;
  }
  return n;
}

uint64_t lcm(uint64_t m, uint64_t n) {
  return (m*n)/gcd(m,n);
}

// vector<vector<uint64_t> > v(n+1, vector<uint64_t>(n+1, 0))
// v[n][k]に組み合わせ数が入る。
void comb(vector<vector <uint64_t> > &v){
  for(uint64_t i = 0;i <v.size(); i++){
    v[i][0]=1;
    v[i][i]=1;
  }
  for(uint64_t k = 1;k <v.size();k++){
    for(uint64_t j = 1;j<k;j++){
      v[k][j]=(v[k-1][j-1]+v[k-1][j]);
    }
  }
}

// 掛け算オーバーフロー判定
bool is_product_overflow(uint64_t a, uint64_t b) {

	uint64_t prod = a * b;
	
	return (prod / b != a);
}

//素因数分解
void primeFactorization(uint64_t a, list<uint64_t> &factors){ //素因数分解を出力するプログラム
  long i,sq;
  if(a%2==0){ //偶数の場合
    factors.push_back(2);
    primeFactorization(a/2,factors); //2で割った値で再帰
    return;
  }
  sq = sqrt(a);
  for(i=3;i<=sq;i+=2){ //3以上√a以下の奇数の場合
	if(a%i==0){
	  factors.push_back(i);
      primeFactorization(a/i,factors); //割れた値で再帰
      return;
	}
  }
  //偶数でも3以上√a以下の奇数の場合でも割り切れない場合
  if(a!=1){ //aが1でないなら、a自身は素数
    factors.push_back(a);
  }
}

// フェルマーの小定理
// mod. m での a の逆元 a^{-1} を計算する
// (a/b)%m = a*movinv(b,m)%m  
int64_t modinv(int64_t a, int64_t m) {
    int64_t b = m, u = 1, v = 0;
    while (b) {
        int64_t t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

// 円周率
// M_PI

// #include <iomanip> // setprecisionを使用するのに必要
// cout << std::fixed << std::setprecision(15) << y << endl;

// 昇順
// priority_queue<int, vector<int>, greater<int> > queue;


signed main() {
  int64_t n,k,p;
  cin >> n >> k;
  list<int64_t> valueList;
  int64_t totalCurrent = 0;
  int64_t totalMax = 0;
  for(int64_t i=0;i<n;i++){
    cin >> p;
    valueList.push_back(p);
    totalCurrent += p;
    int size = valueList.size();
    if(size>k){
      totalCurrent -= valueList.front();
      valueList.pop_front();
    }
    if(totalCurrent>totalMax){
      totalMax = totalCurrent;
    }
  }
  cout << fixed << setprecision(10) << ((double)(totalMax+k))/2.0 << endl;
  return 0;
}