Problem:
Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner stdIn=new Scanner(System.in);
		int N=stdIn.nextInt();
		int K=stdIn.nextInt();
		int zero[]=new int[N/2+10];
		int one[]=new int[N/2+10];
		String S=stdIn.next();
		String h="";
		int z=0,y=0,x=0,ir=0,ir2=0,cou=0,max=0,w=0;
		while(z<N){
			if(S.substring(z,z+1).equals("0")){
				if(ir==1&&zero[0]!=0)
					x++;
				zero[x]++; ir=0;
			}
			else{
				if(ir==0&&one[0]!=0)
					y++;
				one[y]++; ir=1;
			}
			z++;
		}z=0;ir=0;
		while(z<K){
			h=S.substring(z,z+1);
			if(z==0&&ir==0){
				if(h.equals("0")){
					cou+=zero[z];
					cou+=one[z];
				}
				else{
					cou+=one[z];
					z--;
					ir=1;
				}
			}
			else{
				cou+=zero[z];
				cou+=one[z+ir];
			}
			z++;
		}max=cou;
		while(z<x||z<y){
			if(ir==0&&w==0){
				cou-=zero[w];
				cou+=zero[z];
				cou+=one[z];
			}
			else{
				cou-=zero[w];
				cou-=one[w];
				cou+=zero[z];
				cou+=one[z+ir];
			}
			if(max<cou)
				max=cou;
			z++;w++;
		}
		System.out.println(max);
	}
}