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: Runtime Error)

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();
        String s = sc.next();
        List<Integer> kList = new ArrayList<>();
        List<Integer> oneList = new ArrayList<>();
        
        boolean f = true;
        int z = 0;
        int one = 0;
        boolean firstoneFlag = false;
        for(int i = 0; i < n; i++){
            z++;
            char c = s.charAt(i);
            if(i == 0 && c == '1'){
                firstoneFlag = true;
            }
            if(c == '1'){
                if(f){
                    one++;
                }else{
                    f = true;
                    kList.add(z);
                    if(firstoneFlag){
                        firstoneFlag = false;
                    }else{
                        oneList.add(one);
                    }
                    one = 1;
                }
            }else{
                f = false;
            }
        }
        
        if(f){
            oneList.add(one);
        }else{
            kList.add(z);
            oneList.add(0);
        }
        
        long max = 0;
        int index = -1;
        for(int i = k - 1; i < kList.size(); i++){
            long tmp = 0;
            if(index == -1){
                tmp = kList.get(i) + oneList.get(i);
            }else{
                tmp = kList.get(i) + oneList.get(i) - kList.get(index);
            }
            if(max < tmp){
                max = tmp;
            }
            index++;
        }
        
        if(max == 0){
            max = n;
        }
        
        System.out.println(max);
        
        
        
    }
}