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 Scala (2.11.7) (judge verdict: Wrong Answer)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val N, K = sc.nextInt()
    sc.nextLine()
    val S: Array[Char] = sc.nextLine().toCharArray

    var left = 0
    while (left < N && S(left) == '1')
      left += 1

    var k = 0
    var right = left
    while (k < K) {
      while (right < N && S(right) == '0')
        right += 1
      while (right < N && S(right) == '1')
        right += 1
      k += 1
    }

    var max = right - left
    while (left < N && right < N) {
      while (left < N && S(left) == '0')
        left += 1
      while (left < N && S(left) == '1')
        left += 1
      while (right < N && S(right) == '0')
        right += 1
      while (right < N && S(right) == '1')
        right += 1
      max = Math.max(max, right - left + 1)
    }

    println(max)
  }

}