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

    String s2 = "_";
    StringBuilder buf = new StringBuilder();
    buf.append(S);
    buf.append(s2);
    S = buf.toString();

    int[] f0 = new int[N];
    int[] f1 = new int[N];
    int count = 0;
    int k0 = 1;
    int k1 = 0;
    f0[0] = count;

    if(!S.contains("01")){
      System.out.println(S.length()-1);
    } else {
      loop:while(true){
        while(true){
          if(S.charAt(count) == "0".charAt(0) && S.charAt(count+1) == "1".charAt(0)){
            f0[k0] = count+1;
            k0++;
            break;
          }
          if(S.charAt(count) == "0".charAt(0) && S.charAt(count+1) == "_".charAt(0)){
            f0[k0] = count+1;
            k0++;
            break loop;
          }
          count++;
        }
        while(true){
          if(S.charAt(count) == "1".charAt(0) && S.charAt(count+1) == "0".charAt(0)){
            f1[k1] = count+1;
            k1++;
            break;
          }
          if(S.charAt(count) == "1".charAt(0) && S.charAt(count+1) == "_".charAt(0)){
            f1[k1] = count+1;
            k1++;
            break loop;
          }
          count++;
        }
      }
      int max = 0;

      for(int i = 0; i <= k0-K-1; i++){
        System.out.println(f1[K-1+i]);
        System.out.println(f0[i]);
        if(max < f1[K-1+i] - f0[i]){
          max = f1[K-1+i] - f0[i];
        }
      }
      System.out.println(max);
    }
  }
}