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: Time Limit Exceeded)

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).toVector

    var s2 = Vector[(Int, Int, Int)]()
    var previous = s(0)
    var from = 0
    for (i <- 1 to s.size) {
      if (i == s.size) {
        s2 = (previous, from, i - from) +: s2
      } else {
        val which = s(i)
        if (which != previous) {
          s2 = (previous, from, i - from) +: s2
          previous = which
          from = i
        }
      }
    }

    val s3 = s2.reverse
    var counts = Vector[Int]()
    val b = new Breaks
    b.breakable {
      for (i <- s3.indices) {
        counts = (if (s3(i)._1 == 1) s3.slice(i, i + k * 2 + 1).map(_._3).sum else s3.slice(i, i + k * 2).map(_._3).sum) +: counts
        if (i > s3.size - k * 2) b.break()
      }
    }
    counts.max.toString
  }
}