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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define rep(i,j,k) for(int i=j;i<k;i++)
#define all(x) (x).begin(),(x).end()
#define UNIQUE(v) v.erase( unique(v.begin(), v.end()), v.end() );
#define pb push_back
#define mp make_pair
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
const int  COMMAX = 510000;

long long fac[COMMAX], finv[COMMAX], inv[COMMAX];

int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcd(int a,int b){return a*b / gcd(a,b);}
// テーブルを作る前処理
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < COMMAX; i++){
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}

// 二項係数計算
long long COM(int n, int k){
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

struct UnionFind {
  vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
  vector<int> rank;
  int num;
  UnionFind(int N) : par(N) , rank(N) { //最初は全てが根であるとして初期化
    num = N;
    for(int i = 0; i < N; i++){
      par[i] = i;
      rank[i] = 0;
    }
  }

  int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
    if (par[x] == x) return x;
    return par[x] = root(par[x]);
  }

  void unite(int x, int y) { // xとyの木を併合
    int rx = root(x); //xの根をrx
    int ry = root(y); //yの根をry
    if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
    if(rank[rx] < rank[ry]){
      par[rx] = ry;  
    }else{
      par[ry] = rx;
      if(rank[rx] == rank[ry])rank[rx]++;
    }
    num--;
  }

  bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
    return root(x) == root(y);
  }
};

const int MS = 2;
const long long mod[] = {999999937LL, 1000000007LL}, base = 9973;
struct rolling_hash {
    int n;
    vector<long long> hs[MS], pw[MS];
    rolling_hash(){}
    rolling_hash(const string &s) {
        n = s.size();
        for (int i = 0; i < MS; i++) {
            hs[i].assign(n+1,0);
            pw[i].assign(n+1,0);
            hs[i][0] = 0;
            pw[i][0] = 1;
            for (int j = 0; j < n; j++) {
                pw[i][j+1] = pw[i][j]*base%mod[i];
                hs[i][j+1] = (hs[i][j]*base+s[j])%mod[i];
            }
        }
    }

    long long hash(int l, int r, int i) { return ((hs[i][r]-hs[i][l]*pw[i][r-l])%mod[i]+mod[i])%mod[i]; }

    bool match(int l1, int r1, int l2, int r2) {
        bool ret = 1;
        for (int i = 0; i < MS; i++) ret &= hash(l1,r1,i)==hash(l2,r2,i);
        return ret;
    }

    bool match(int l, int r, long long h[]) {
        bool ret = 1;
        for (int i = 0; i < MS; i++) ret &= hash(l,r,i)==h[i];
        return ret;
    }
};

int main(){
  int n,k;
  cin >> n >> k;
  vector<double> v(n);
  vector<double> v1(n);
  vector<LL>v2(200000);
  v2[0]  = 1;
  rep(i,1,200000)v2[i] = v2[i-1] + i + 1;
  rep(i,0,n){
    double a;
    cin >> a;
    v[i] = v2[a-1] / a;
  }
  rep(i,1,n){
    v1[i] = v1[i-1] + v[i];
  }
  double res = 0;
  rep(i,0,n-k){
    res = max(res, v1[i+k] - v1[i]);
  }
cout << res << endl;
  return 0;
}