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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        String[] tmp = br.readLine().split(" ");
        int n = Integer.parseInt(tmp[0]);
        int k = Integer.parseInt(tmp[1]);
        char[] s = br.readLine().toCharArray();

        List<Integer> list = new ArrayList<>();
        boolean b = true;
        int i;
        for(i = 0; i < n; i++) {
            if(b != (s[i] == '1')) {
                list.add(i);
                b = !b;
            }
        }
        list.add(i);
        if(list.size() % 2 == 0) list.add(i);

        int ans = 0;
        int j = 0;
        do {
            ans = Math.max(ans, list.get(Math.min(j + 2 * k, list.size() - 1)) - (j == 0? 0 : list.get(j - 1)));
            j += 2 * k;
        }while(j < list.size());

        System.out.println(ans);
    }
}