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.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main implements Runnable {

    private static int MOD = 1_000_000_007;

    public static void main(String[] args) {
        // Run with 32MB stack
        Thread thread = new Thread(null, new Main(), "", 32 * 1024 * 1024);
        thread.start();
    }

    @Override
    public void run() {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }

    static void solve(Scanner scanner) {
        int[] a = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
        int N = a[0], K = a[1];
        String S = scanner.nextLine();
        if (!S.contains("0") || !S.contains("1")) {
            System.out.println(1);
            return;
        }
        List<Integer> starts = new ArrayList<>();
        List<Integer> ends = new ArrayList<>();
        boolean one = false;
        for (int i = 0; i < N; i++) {
            if (S.charAt(i) == '1') {
                if (!one) {
                    one = true;
                    starts.add(i);
                }
            } else {
                if (one) {
                    ends.add(i - 1);
                    one = false;
                }
            }
        }
        if (one) {
            ends.add(N - 1);
        }

        int max = 0;
        for (int i = 0; i < starts.size(); i++) {
            int endIdx = i + K >= ends.size() ? ends.size() - 1 : i + K;
            max = Math.max(max, ends.get(endIdx) - starts.get(i) + 1);
        }

        int endIdx = K - 1 >= ends.size() ? ends.size() - 1 : K - 1;
        max = Math.max(max, ends.get(endIdx) + 1);
        int startIdx = Math.max(starts.size() - K, 0);
        max = Math.max(max, N - starts.get(startIdx));
        System.out.println(max);
    }

}