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: Wrong Answer)

import java.util.*;

public class Main {
	Scanner sc = new Scanner(System.in);

	void doIt() {
		int n = sc.nextInt();
		int k = sc.nextInt();
		String s = sc.next();
		int max = 0;
		char b = '1';
		int len = 0;
		int[] l = new int[1000000];
		l[len] = 0;
		len++;
		for(int i = 0; i < n; i++) {
			char c = s.charAt(i);
			if(b != c) {
				l[len] = i;
				len++;
				b = c;
			}
		}
		l[len] = n;
		len++;
		int idx = 0, d = 2 * k + 1;
		//System.out.println(Arrays.toString(l.toArray()));
		while(idx + d < len) {
			int x = l[idx + d] - l[idx];
			if(max < x) max = x;
			idx += 2;
			//System.out.println(max);
		}
		if(max == 0) {
			if(!(k == 0 && s.contains("1"))) max = n;
		}
		System.out.println(max);
		
	}
	
	public static void main(String[] args) {
		new Main().doIt();
	}
}