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.*;

class Block {
  int count;
  boolean reverse;
  
  Block(int count, boolean reverse) {
    this.count = count;
    this.reverse = reverse;
  }
}

public class Main {
  private static int solve(ArrayList<Block> blocks, int k) {
    int sum = blocks.get(0).count;
    int maxSum = sum;
    for (int i = 1; i+1 < blocks.size(); i+=2) {
      sum += blocks.get(i).count;    // flipped
      sum += blocks.get(i+1).count;  // originally reversed
      if (i - 2 * k - 1 >= 0) {
        sum -= blocks.get(i - 2 * k).count;
        sum -= blocks.get(i - 2 * k - 1).count;
      }
      maxSum = Math.max(sum, maxSum);
    }
    return maxSum;
  }
  
  public static void main(String[] args) {
    Scanner s = new Scanner(System.in);
    
    int n = s.nextInt();
    int k = s.nextInt();
    String line = s.next();
    ArrayList<Block> blocks = new ArrayList<Block>();
    
    
    
    int count = 0;
    char[] cs = line.toCharArray();
    count = 1;
    boolean lastReverse = cs[0] == '1';
    if (!lastReverse) {
      blocks.add(new Block(0, true));
    }
    
    for (int i = 1; i < cs.length; i++) {
      boolean currentReverse = cs[i] == '1';
      if (lastReverse == currentReverse) {
        count++;
      } else {
        blocks.add(new Block(count, lastReverse));
        count = 1;
        lastReverse = currentReverse;
      }
    }
    blocks.add(new Block(count, lastReverse));
    if (!lastReverse) {
      blocks.add(new Block(0, true));
    }
    blocks.add(new Block(0, false));
    blocks.add(new Block(0, true));
    
    System.out.println(solve(blocks, k));
  }
}