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: Wrong Answer)

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

//マクロ
//forループ関係
//引数は、(ループ内変数,動く範囲)か(ループ内変数,始めの数,終わりの数)、のどちらか
//Dがついてないものはループ変数は1ずつインクリメントされ、Dがついてるものはループ変数は1ずつデクリメントされる
#define REP(i,n) for(lol i=0;i<lol(n);i++)
#define REPD(i,n) for(lol i=n-1;i>=0;i--)
#define FOR(i,a,b) for(lol i=a;i<=lol(b);i++)
#define FORD(i,a,b) for(lol i=a;i>=lol(b);i--)
#define VL vector<lol>
//xにはvectorなどのコンテナ
#define ALL(x) x.begin(),x.end() //sortなどの引数を省略したい
#define SIZE(x) lol(x.size()) //sizeをsize_tからllに直しておく
//定数
#define PI 3.1415926535897932385 //pi
#define INF 1000000000000 //10^12:極めて大きい値,∞
#define MOD 1000000007 //10^9+7:合同式の法
#define MAXR 100000 //10^5:配列の最大のrange(素数列挙などで使用)
//最大値最小値
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int main()
{
    double result{0};
    lol n; cin >> n;
    lol k; cin >> k;
    VL p(n);
    REP(i, n)
    {
        cin >> p[i];
    }
    FOR(i, 0, k-1)
    {
        double nume = (p[i]*(p[i]+1))/2;
        result += nume/p[i];
    }
    double MA = result;
    FOR(i, 1, n-k)
    {
        double nume1 = (p[i-1]*(p[i-1]+1))/2;
        MA -= nume1/p[i-1];
        double nume2 = (p[k+i-1]*(p[k+i-1]+1))/2;
        MA += nume2/p[k+i-1];
        result = max(result, MA);
    }

    cout << result << endl;
    return 0;
}