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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.*;
import java.util.*;
class Main{
	static final long MOD = 1_000_000_007; // 10^9+7
    static final int MAX = 2_147_483_646; // intMax 
    static final int INF = 1_000_000_000; // 10^9  
    public static void main(String[] args) throws Exception {
		hayami saori = new hayami();
		int n = saori.saori_hayami();
		int k = saori.saori_hayami();
		
		int[] p = new int[n];
		
		int sum = 0;
		int max = 0;
		for(int i = 0 ; i < n ; i++ ) {
			p[i] = saori.saori_hayami();
			sum += p[i];
			if ( i >= k ) {
				sum -= p[i-k];
			}
			if ( sum > max ) {
				max = sum;
			}
		}
		System.out.println( ((double)max + (double)k)/2f );
		
		saori.close();
    }
}

class hayami implements Closeable {
	private final InputStream in = System.in;
	private final byte[] hayami = new byte[1024];
	private int Hayami = 0;
	private int saori = 0;
	private boolean HayamiSaori() {
		if (Hayami < saori) {
			return true;
		}else{
			Hayami = 0;
			try {
				saori = in.read(hayami);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (saori <= 0) {
				return false;
			}
		}
		return true;
	}
	private int SaoriHayami() { 
		if (HayamiSaori()) {
            return hayami[Hayami++];
         }else{
             return -1;
         }
	}
	private static boolean hayami_saori(int hayami) { 
		return 33 <= hayami && hayami <= 126;
	}
	public boolean hayamisaori() { 
		while(HayamiSaori() && !hayami_saori(hayami[Hayami])) Hayami++; return HayamiSaori();
	}
	public String nextHayami() {
		if (!hayamisaori()) throw new NoSuchElementException();
		StringBuilder hayamin = new StringBuilder();
		int saori = SaoriHayami();
		while(hayami_saori(saori)) {
			hayamin.appendCodePoint(saori);
			saori = SaoriHayami();
		}
		return hayamin.toString();
	}
	public long saorihayami() {//nextLong
		if (!hayamisaori()) throw new NoSuchElementException();
		long hayami = 0;
		boolean misao = false;
		int saori = SaoriHayami();
		if (saori == '-') {
			misao = true;
			saori = SaoriHayami();
		}
		if (saori < '0' || '9' < saori) {
			throw new NumberFormatException();
		}
		while(true){
			if ('0' <= saori && saori <= '9') {
				hayami *= 10;
				hayami += saori - '0';
			}else if(saori == -1 || !hayami_saori(saori)){
				return misao ? -hayami : hayami;
			}else{
				throw new NumberFormatException();
			}
			saori = SaoriHayami();
		}
	}
	public int saori_hayami() {//nextInt
		long hayami = saorihayami();
		if (hayami < Integer.MIN_VALUE || hayami > Integer.MAX_VALUE) throw new NumberFormatException();
		return (int) hayami;
	}
	public double Hayamin() { //nextDouble
		return Double.parseDouble(nextHayami());
	}
	public void close() {
		try {
			in.close();
		} catch (IOException e) {
		}
    }
    
}