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.Scanner;
import java.util.TreeSet;

public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		String s = sc.next();
		char[] arr = s.toCharArray();
		sc.close();

		int ans = 0;
		int cnt1 = 0;
		int cnt0 = 0;
		int siji = 0;
		TreeSet<Integer> set = new TreeSet<Integer>();
		set.add(-1);
		for (int i = 0; i < n; i++) {
			if (arr[i] == '1') {
				cnt0 = 0;
				if (cnt1 == 0) {
					if (siji > k) {
						set.pollFirst();
					}
					Integer idx = set.first();
					ans = Math.max(ans, i - idx);
					set.add(i);
				}
				cnt1++;
			} else {
				cnt1 = 0;
				if (cnt0 == 0) {
					siji++;
				}
				cnt0++;
			}
		}
		if (siji > k) {
			set.pollFirst();
		}
		Integer idx = Math.max(set.first(), 0);
		ans = Math.max(ans, n - idx);
		System.out.println(ans);
	}
}