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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.PrintStream;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws Exception{
		new Main(new Scanner(System.in), System.out, System.err).exec();
	}
	public final Scanner sc;
	public final PrintStream out, err;
	public Main(final Scanner sc, final PrintStream out, final PrintStream err) {
		this.sc = sc;this.out = out;this.err = err;
	}
	static public final long pair(int a, int b) {return (((long)a) << 32) + b;}
	static public final int getPair1(long v) {return (int)(v >> 32);}
	static public final int getPari2(long v) {return (int)(v & 0xFFFFFFFF);}

	static public final long quad(long a, long b, long c, long d) {
		return ((a&0xFFFF)<< 48) + ((b&0xFFFF)<< 32) + ((c&0xFFFF)<< 16) +(d&0xFFFF);}
	static public final int getQuad1(long v) {return (int)( v >> 48          );}
	static public final int getQuad2(long v) {return (int)((v >> 32) & 0xFFFF);}
	static public final int getQuad3(long v) {return (int)((v >> 16) & 0xFFFF);}
	static public final int getQuad4(long v) {return (int)( v        & 0xFFFF);}

	static public final long record(int e, int m, int i) {
		return quad(-1, e, m, i);
	}
	static public final int re(long record) {
		return getQuad2(record);
	}
	static public final int rm(long record) {
		return getQuad3(record);
	}
	static public final int ri(long record) {
		return getQuad4(record);
	}

	//------------------------------------------

	private int p;
	private int[] inv;
	private int[] fact;
	private int[] finv;
	public final int mod(long v) {
		v %= p;
		return (int)(v < 0 ? v+p : v);
	}
	public final int mul(long a, long b) {
		return mod(a*b);
	}
	public final void initFact(double limit, double p) {
		initFact((int)limit, (int)p);
	}
	public final void initFact(int limit, int p) {
		this.p = p;
		this.inv = new int[limit+1];
		this.fact = new int[limit+1];
		this.finv = new int[limit+1];
		fact[1] = finv[1] = inv[0] = inv[1] = 1;
		for(int i=2;i<=limit;i++) {
			int a = p / i;
			int b = p % i;
			inv[i]  = mul(-a, inv[b]);
			fact[i] = mul(fact[i-1], i);
			finv[i] = mul(finv[i-1], inv[i]);
		}
	}
	public final int comb(int m, int n) {
		return mul(mul(fact[m], finv[n]), finv[m-n]);
	}
	//------------------------------------------
	static long lcm (long a, long b) {
		long temp;
		long c = a;
		c *= b;
		while((temp = a%b)!=0) {
			a = b;
			b = temp;
		}
		return (long)(c/b);
	}

	//------------------------------------------



	public void exec() throws Exception{
		final int N = sc.nextInt();
		final int K = sc.nextInt();
		final int[] p = new int[N];
		for(int i=0;i<N;i++) {
			p[i] = sc.nextInt();
		}

		int max = 0;
		int sum = 0;
		for(int i=0;i<K;i++) {
			sum += p[i];
		}
		max = sum;
		for(int i=K;i<N;i++) {
			sum = sum - p[i-K] + p[i];
			max = Math.max(sum, max);
		}

		out.println((max+K)/2.0);
	}
}