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 Java7 (OpenJDK 1.7.0) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main{

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), k = sc.nextInt();
		String s = sc.next();
		sc.close();
		int ans = 0;
		int j = 0;
		int tmp = 0;
		int cnt = 0;
		for(int i = 0; i < n; ++i) {
			if(s.charAt(i) == '1') 
				tmp++;
			else {
				if(i != 0) {
					if(s.charAt(i - 1) == '1')
						++cnt;
				}
				else ++cnt;
				if(cnt == 1)j = i;
				if(cnt <= k)++tmp;
			}
			if(cnt > k) {
				ans = Math.max(ans,  tmp);
				cnt = 0;
				tmp = 0;
				i = j;
			}
			else {
				if(i == n - 1) 
					ans = Math.max(ans,  tmp);
			}
		}
		
		System.out.println(ans);
		
	}

}