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>
#define FOR(i, m, n) for (int i = m; i < (n); i++)
#define RFOR(i, m, n) for (int i = (m - 1); i >= 0; i--)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) RFOR(i, n, 0)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const ll INF = 1000000000000000000LL;

struct mint {
  ll x;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return (*this) *= a.inv(); }
  mint operator/(const mint a) const {
    mint res(*this);
    return res /= a;
  }
};

mint factorial(ll n) {
  mint answer = 1;
  while (n > 1) {
    answer *= n;
    n--;
  }
  return answer;
}

mint combination(ll n, ll r) {
  return factorial(n) / (factorial(r) * factorial(n - r));
}

mint permutation(ll n, ll r) { return factorial(n) / factorial(n - r); }

// setprecision(8) 桁数指定
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }

ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }

template <typename T>
void remove(std::vector<T>& vector, unsigned int index) {
  vector.erase(vector.begin() + index);
}

map<ll, ll> prime_factor(ll n) {
  map<ll, ll> ret;
  for (ll i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if (n != 1) ret[n] = 1;
  return ret;
}
// for (auto p : prime_factor(n)) {
//	while (p.second--) {
//		cout << p.first << endl;
//	}
//}

vector<ll> divisor(ll n) {
  vector<ll> ret;
  for (ll i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}

bool compare_by_b(pair<ll, ll> a, pair<ll, ll> b) {
  if (a.second != b.second) {
    return a.second < b.second;
  } else {
    return a.first < b.first;
  }
}

vector<ll> Eratosthenes(const ll N) {
  bool is_prime[1000000 + 1];
  vector<ll> P;
  for (ll i = 0; i <= N; i++) {
    is_prime[i] = true;
  }
  for (ll i = 2; i <= N; i++) {
    if (is_prime[i]) {
      for (ll j = 2 * i; j <= N; j += i) {
        is_prime[j] = false;
      }
      P.emplace_back(i);
    }
  }
  return P;
}

typedef vector<int> ivec;
typedef vector<string> svec;
typedef vector<ll> lvec;

int main() {
  int n, k;
  cin >> n >> k;
  vector<int> p(n);
  REP(i, n) { cin >> p[i]; }
  double ans = 0;
  vector<double> pp(n);

  REP(j, n) {
    int p_j = p[j];
    double a_j = 1 / (double)2 * p_j * (p_j + 1) / p_j;
    pp[j] = a_j;
  }

if (n == 1) {
  double a = 1 / (double)2 * p[0] * (p[0] + 1) / p[0];
  cout << a << endl;
  return 0;
}

  double sum = 0;
  REP(i, k) { sum += pp[i]; }

  REP(i, n - k) {
    sum = sum - pp[i] + pp[i + k];
    ans = max(ans, sum);
  }

  cout << fixed << setprecision(10) << ans << endl;
  return 0;
}