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) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		String s = sc.next();
		String[] Num = s.split("");
		int[] num = new int[n];
		for(int i = 0; i < n; i++) {
			num[i] = Integer.parseInt(Num[i]);
		}
		long ans = 0;
		int change = 0;
		int j = 1;
		while(j < n) {
			if(num[j-1] + num[j] == 1) {
				if(num[j-1] == num[j+1]) {
					change ++;
					j += 2;
				}else {
					change ++;
					j += 1;
				}
			}else {
				j ++;
			}
		}
		if(change <= k && num[0] == 1) {
			ans = n;
		}else if(change < k && num[0] == 0) {
			ans = n;
			//System.out.println("jack");
		}else if(num[0] == 1){
			int[] tng = new int[change*2 + 2];
			int ind = 0;
			tng[0] ++;
			for(int i = 1; i < n; i++) {
				if(num[i-1] + num[i] != 1) {
					tng[ind] ++;
				}else {
					ind ++;
					tng[ind] ++;
				}
			}
			int L = 0;

			while(L + 2*k < tng.length ) {
				int tmp = 0;
				for(int i = L; i <= L + 2*k; i++) {
					tmp += tng[i];
				}
				L += 2;
				ans = Math.max(ans, tmp);
			}
		}else {
			int[] tng = new int[change*2 + 2];
			int ind = 0;
			tng[0] ++;
			for(int i = 1; i < n; i++) {
				if(num[i-1] + num[i] != 1) {
					tng[ind] ++;
				}else {
					ind ++;
					tng[ind] ++;
				}
			}

			int L = 0;
			for(int i = L; i < L + 2*k; i++) {
				ans += tng[i];
			}
			L++;

			while(L + 2*k < tng.length) {
				int tmp = 0;
				for(int i = L; i <= L + 2*k; i++) {
					tmp += tng[i];
				}
				L += 2;
				ans = Math.max(ans, tmp);
			}
		}
		System.out.println(ans);

		sc.close();
	}

}