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)

import scala.util.control.Breaks

object Main {
  def main(args: Array[String]): Unit = {
    val lines = io.Source.stdin.getLines().toSeq
    println(solve(lines))
  }

  def solve(lines: Seq[String]): String = {
    val Array(n, k) = lines.head.split(" ").map(_.toInt)
    val s = lines.tail.head.split("").filter(_.length > 0).map(_.toInt).toIndexedSeq

    var s2 = IndexedSeq[(Int, Int, Int)]()
    val b1 = new Breaks
    b1.breakable {
      var i = 0
      while (true) {
        val b2 = new Breaks
        b2.breakable {
          for (j <- i + 1 to s.size) {
            if (j == s.size || s(i) != s(j)) {
              s2 = s2 :+ (s(i), i, j - i)
              i = j
              b2.break()
            }
          }
        }
        if (i == s.size) b1.break()
      }
    }

    var counts = Seq[Int]()
    for (i <- 0 until s2.size - k * 2) {
      counts = counts :+ (if (s2(i)._1 == 1) s2.slice(i, i + k * 2 + 1).map(_._3).sum else s2.slice(i, i + k * 2).map(_._3).sum)
    }
    if (counts.isEmpty) s.count(_ == 1).toString else counts.max.toString
  }
}