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>
#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower

#define int long long

// loop
#define FOR(i, a, b) for (int i = (a); i < (b); i++) 	// a ~ b-1	(ascending)
#define REP(i, n) FOR(i, 0, n)							// 0 ~ n-1
#define NREP(i, n) FOR(i, 1, n + 1)						// 1 ~ n
#define RFOR(i, a, b) for (int i = (a); i >= (b); i--)	// a ~ b	(descending)
#define RREP(i, n) RFOR(i, n, 0)						// n ~ 0
#define RNREP(i, n) RFOR(i, n, 1)						// n ~ 1
 
// container operation
#define all(v) v.begin(), v.end()
#define EACH(i, c) for (auto i = (c).begin(); i != (c).end(); i++)
#define ASORT(c) std::sort((c).begin(), (c).end())
#define DSORT(c) std::sort((c).begin(), (c).end(), std::greater<typeof((c).front())>())
#define SIZE(x) ((int)(x).size())
// union of two containers
template <typename T>
void container_union(T &c1, T &c2, T &c3) {
  std::vector<int> vec;
  std::set_union(c1.begin(), c1.end(),
                 c2.begin(), c2.end(),
                 std::back_inserter(vec));
  T tmp(vec.begin(), vec.end());
  c3 = tmp;
}

// chmax, chmin
template <typename T>
bool chmax(T &a, const T& b) {
  if (a < b) {
    a = b;  // aをbで更新
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T &a, const T& b) {
  if (a > b) {
    a = b;  // aをbで更新
    return true;
  }
  return false;
}

// grid-search
std::vector<int> dx = {1, -1, 0,  0};
std::vector<int> dy = {0,  0, 1, -1};
 
// debug
#define check(x) std::cout << #x << " = " << x << '\n'
 
// print
#define cout(x) std::cout << x << '\n'
 
// type alias
using VI = std::vector<int>;
using VII = std::vector<VI>;
using VB = std::vector<bool>;
using VBB = std::vector<VB>;
using VS = std::vector<std::string>;
using PII = std::pair<int, int>;

// yes/no
std::string yes = "Yes";
std::string no = "No";
std::string YES = "YES";
std::string NO = "NO";

void solve();

signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  solve();
  return 0;
}

void solve() {
  int n, k;
  std::cin >> n >> k;
  VI p(n);
  std::vector<int> e(n);
  REP(i, n) {
    std::cin >> p.at(i);
    e.at(i) = p.at(i) + 1;
  }

  int max = 0;
  int sum = 0;
  REP(i, n - k + 1) {
    if (i == 0) {
      REP(j, k) {
        sum += e.at(j);
      }
    } else {
      sum -= e.at(i-1);
      sum += e.at(i+k-1);
    }
    chmax(max, sum);
  }
  cout((long double)max / (long double)2);
}