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.*;

public class Main{
  public static void main(String[]args){
    Scanner sc = new Scanner(System.in);
    String[] S = sc.nextLine().split(" ");
    int N = Integer.parseInt(S[0]);
    int K = Integer.parseInt(S[1]);
    String T = sc.nextLine();
    int a = 0, b = 0, count = 0, max = 0, now = 0;
    boolean check = true;
    while(a < N){
      while(count <= K && b < N){
        now++;
        if(b < N-1){
          if(T.charAt(b) == '0' && T.charAt(b+1) == '1'){
            count++;
          }
        }else{
          if(T.charAt(b) == '0'){
            count++;
            now--;
          }
        }
        b++;
        max = max < now ? now : max;
      }
      
      now--;
      if(a < N-1){
          if(T.charAt(a) == '0' && T.charAt(a+1) == '1'){
            count--;
          }
        }else{
          if(T.charAt(a) == '0'){
            count--;
          }
        }
      a++;
      if(b == N-1 && count < K){
        break;
      }
    }   
    System.out.println(max);
  }
}