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: Time Limit Exceeded)

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

public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        int K = sc.nextInt();
        sc.nextLine();
        String S = sc.nextLine();
        sc.close();

        int pos = 0;
        List<Integer> num = new ArrayList<Integer>();
        List<Character> flag = new ArrayList<Character>();
        while (pos < N) {
            int count = 0;
            char curObj = S.charAt(pos);
            while (S.charAt(pos) == curObj) {
                count++;
                if (pos < N - 1) {
                    pos++;
                } else {
                    break;
                }
            }
            num.add(count);
            flag.add(curObj);

            if (S.charAt(pos) != curObj) {
                continue;
            } else if (pos == N - 1){
                break;
            }
        }
        // System.out.println(num);
        // System.out.println(flag);

        List<Integer> sumLog = new ArrayList<Integer>();
        List<Integer> sumFlag = new ArrayList<Integer>();
        int max_cur = 0;
        for (int i = 0; i < num.size(); i++) {
            // System.out.println("Round" + i);
            if (flag.get(i) == '0') {
                // System.out.println(flag.get(i));
                // System.out.println(sumLog);
                // System.out.println(sumFlag);
                for (int m = 0; m < sumLog.size(); m++) {
                    if (sumFlag.get(m) == 0) {
                        sumLog.remove(m);
                        sumFlag.remove(m);
                        m--;
                        // System.out.println("inside of first loop");
                    }
                }
                // System.out.println("first loop");
            }

            for (int j = 0; j < sumLog.size(); j++) {
                sumLog.set(j, sumLog.get(j) + num.get(i));
                if (flag.get(i) == '0') {
                    sumFlag.set(j, sumFlag.get(j) - 1);
                }
                // System.out.println(sumLog);
                // System.out.println(sumFlag);
                // System.out.println("second loop");
            }

            sumLog.add(num.get(i));
            if (flag.get(i) == '0') {
                sumFlag.add(K - 1);
            } else {
                sumFlag.add(K);
            }

            for (int k = 0; k < sumLog.size(); k++) {
                if (max_cur < sumLog.get(k)) {
                    max_cur = sumLog.get(k);
                }
            }
            // System.out.println(max_cur);
            // System.out.println(sumLog);
            // System.out.println(sumFlag);
            // System.out.println("================");
        }
        System.out.println(max_cur);
    }
}