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 extends App {

  val line1 = io.StdIn.readLine.split(' ').map(_.toInt)
  val s = io.StdIn.readLine

  val N = line1(0)
  val K = line1(1)

  var previous = '-'

  var seq = Seq.empty[Int]
  var accum = 0

  for (i <- 0 until s.size) {
    s(i) match {
      case ss if ss == previous => accum += 1
      case ss if ss != previous && accum > 0=> {
        seq = accum +: seq
        accum = 1
      }
      case _ => accum += 1
    }
    previous = s(i)
  }

  if (accum > 0)
    seq = accum +: seq

  val num = if (s.head == '0') 2*K else 2*K+1
  val ans = seq.sliding(num).map(_.sum).max

  println(ans)

}