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: Accepted)

#define ll long long int
#include <bits/stdc++.h>
//#include <fstream>
using namespace std;
 
#define         ff               first
#define         ss               second
#define         mp               make_pair
#define         pb               push_back
#define         pf               push_front
#define         vi               vector<ll>
#define         pii              pair<ll,ll>
#define         qi               queue<ll>
#define         mem(a,b)         memset(a,b,sizeof(a))
#define         print(a)         cout<<a<<'\n'
#define         sorted(arr)      sort(arr.begin(),arr.end())
#define         gcd(a,b)         gcd((a),(b))
#define         lcm(a,b)         ((a)*(b))/gcd((a),(b))
#define         all(v)           v.begin(),v.end()
#define         read(a)          scanf("%d", &a)
#define         f(a,n)           for(ll i=a;i<n;i++)
#define         in2(n,m,arr)     for(ll i=0;i<n;i++){for(ll j=0;j<m;j++)cin>>arr[i][j];}
#define         debug(a)         cerr<<a<<" ";
#define         debug1(a)        cerr<<a<<'\n'
 
/*typedef tree<int, null_type, less<int>, rb_tree_tag , tree_order_statistics_node_update> Set;*/
 
 
const unsigned ll mod = 1e9 + 7 ;
const ll inf = (ll)1e18 + 2;
const ll N = (ll)1e6 + 2;
//
//template<typename... T>
//void write(T&... args)
//{
//    ((cout << args << ' '), ...);
//}
//ll power ( ll x , ll y)
//{
//    ll res = 1;
//    while (y > 0)
//    {
//        if (y & 1)
//            res = (res * x) ;
//        y = y >> 1;
//        x = (x * x) ;
//    }
//    return res  ;
//}
//
//struct node{
//    ll x , y , z;
//};
//map<pair<ll, ll>,ll> c;
//vector<set<ll>> tree;
//vector<ll> ar ;
//set<ll> s;
//ll get(ll x)
//{
//    if(x == ar[x])
//        return x;
//    return ar[x] = get(ar[x]);
//}
//void merge(ll a ,ll b)
//{
//    a = get(a);
//    b = get(b);
//    if(a == b)
//    {
//        return ;
//    }
//    ar[b] = ar[a];
// }
//
//ll sev[100006];
//vector<ll> pri;
//void seive()
//{
//    for(ll i = 2 ; i < 100006 ; i ++)
//    {
//        if(!sev[i])
//        {
//            for(ll j = i ; j < 100006 ; j += i)
//            {
//                if(!sev[j])
//                    sev[j] = i;
//            }
//        }
//        if(sev[i] == i)
//            pri.pb(i);
//    }
//}
void solve()
{
    ll n , k;
    cin >> n >> k;
    double arr[n];
    for(ll i = 0 ; i < n ; i ++)
    {
        cin >> arr[i];
        arr[i] = (arr[i] + 1) / 2;
    }
    double sum = 0;
    for(ll i = 0 ; i < k ;i ++)
        sum += arr[i];
    ll l = 0 ;double ans = 0;
    ans = max(ans , sum);
    for(ll i = k ; i < n ; i ++)
        sum -= arr[l] , sum += arr[i] ,ans = max(ans , sum) , l ++;
    cout << fixed << setprecision(10) << ans << '\n';
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll t ;
    t = 1;
    //cin >> t;
    //seive();
    while(t --)
    {
        solve();
    }
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}