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 Python (3.4.3) (judge verdict: Accepted)

import math
import queue
from collections import defaultdict

def readInt():
	return int(input())
def readInts():
	return list(map(int, input().split()))
def readChar():
	return input()
def readChars():
	return input().split()
def factorization(n):
	res = []
	if n%2==0:
		res.append(2)
	for i in range(3,math.floor(n//2)+1,2):
		if n%i==0:
			c = 0
			for j in res:
				if i%j==0:
					c=1
			if c==0:
				res.append(i)
	return res
def fact2(n):
	p = factorization(n)
	res = []
	for i in p:
		c=0
		z=n
		while 1:
			if z%i==0:
				c+=1
				z/=i
			else:
				break
		res.append([i,c])
	return res
def fact(n):#階乗
	ans = 1
	m=n
	for _i in range(n-1):
		ans*=m
		m-=1
	return ans
def comb(n,r):#コンビネーション
	if n<r:
		return 0
	l = min(r,n-r)
	m=n
	u=1
	for _i in range(l):
		u*=m
		m-=1
	return u//fact(l)
def combmod(n,r,mod):
	return (fact(n)/fact(n-r)*pow(fact(r),mod-2,mod))%mod
def printQueue(q):
	r=qb
	ans=[0]*r.qsize()
	for i in range(r.qsize()-1,-1,-1):
		ans[i] = r.get()
	print(ans)
def dq():
	return queue.deque()
class UnionFind():
	def __init__(self, n):
		self.n = n
		self.parents = [-1]*n

	def find(self, x): # root
		if self.parents[x]<0:
			return x
		else:
			self.parents[x] = self.find(self.parents[x])
			return self.parents[x]

	def union(self,x,y):
		x = self.find(x)
		y = self.find(y)

		if x==y:
			return

		if self.parents[x]>self.parents[y]:
			x,y = y,x

		self.parents[x]+=self.parents[y]
		self.parents[y]=x

	def size(self,x):
		return -1*self.parents[self.find(x)]

	def same(self,x,y):
		return self.find(x)==self.find(y)

	def members(self,x): # much time
		root = self.find(x)
		return [i for i in range(self.n) if self.find(i)==root]

	def roots(self):
		return [i for i,x in enumerate(self.parents) if x<0]

	def group_count(self):
		return len(self.roots())

	def all_group_members(self):
		return {r: self.members(r) for r in self.roots()} # 1~n
def bitArr(n):#ビット全探索
	x = 1
	zero = "0"*n
	ans = []
	ans.append([0]*n)
	for i in range(2**n-1):
		ans.append(list(map(lambda x:int(x),list((zero+bin(x)[2:])[-1*n:]))))
		x+=1
	return ans;
def arrsSum(a1,a2):
	for i in range(len(a1)):
		a1[i]+=a2[i]
	return a1
def maxValue(a,b,v):
	v2 = v
	for i in range(v2,-1,-1):
		for j in range(v2//a+1): #j:aの個数
			k = i-a*j
			if k%b==0:
				return i
	return -1

n,k = readInts()
p = readInts()
#print(p)
maxValue = sum(p[0:k])
be = maxValue
memo = p[0:k]
ans = 0
for i in range(1,n-k+1):
	be+=-p[i-1]+p[i+k-1]
	if maxValue<be:
		maxValue = sum(p[i:i+k])
		memo = p[i:i+k]

for i in range(k):
	ans+=(memo[i]+1)/2

print(ans)