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: Accepted)

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

public class Main {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int N = Integer.parseInt(sc.next());
        int K = Integer.parseInt(sc.next());
        String S = sc.next();
        sc.close();

        int[] Sn = new int[N];
        List<Integer> LengthList = new ArrayList<>();

        //連続する0、1の数を数える、0を-1で表す
        for (int i=0; i < N; i++) {

            Sn[i] = Integer.parseInt(String.valueOf(S.charAt(i)));

            if (i == 0 || Sn[i-1] != Sn[i]) {

                if(Sn[i] == 1) {

                    LengthList.add(1);
                } else {

                    LengthList.add(-1);
                }
            } else {

                if(Sn[i] == 1) {

                    LengthList.set(LengthList.size()-1, LengthList.get(LengthList.size()-1) + 1);
                } else {

                    LengthList.set(LengthList.size()-1, LengthList.get(LengthList.size()-1) - 1);
                }
            }

        }

        int sum = 0;
        int ans = 0;
        int hitZeroArray = 0;
        int i = 0;
        int j = 0;

        while ( i < LengthList.size()) {

            if (hitZeroArray < K) {

                if (LengthList.get(i) > 0) {

                    sum += LengthList.get(i);
                    i++;

                    if (i == LengthList.size()) {

                        if (ans < sum) {

                            ans = sum;
                        }
                    }
                } else {

                    sum -= LengthList.get(i);
                    hitZeroArray++;
                    i++;

                    if (i == LengthList.size()) {

                        if (ans < sum) {

                            ans = sum;
                        }
                    }
                }
            }  else {

                sum += LengthList.get(i);
                i++;

                if (ans < sum) {

                    ans = sum;
                }

                while(j < LengthList.size()) {


                    sum -= Math.abs(LengthList.get(j));

                    if (LengthList.get(j) < 0 ) {

                        hitZeroArray--;
                        j++;
                        break;
                    }

                    j++;
                }
            }
        }

        System.out.println(ans);
    }

}