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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

fun main(args: Array<String>) {
    val (n, k) = readLine()!!.split(' ').map { it.toInt() }
    val s = readLine()!!
    val xs = split(s, k)

    val r = (0 until xs.size - (k * 2) step 2).map {
        val x = xs.slice(it..(it + (k * 2))).sum()
        x
    }.max()!!

    println(r)
}

private fun split(s: String, k: Int): List<Int> {
    val xs = IntArray(k * 2) { 0 }.toMutableList()

    if (s[0] == '0') {
        xs.add(0)
    }

    if (s.length == 1) {
        xs.add(1)
        return xs
    }

    var x = 1
    var c = s[0]
    s.substring(1).forEach {
        if (it != c) {
            xs.add(x)
            x = 0
        }
        x++
        c = it
    }

    xs.add(x)

    if (s.last() == '0') {
        xs.add(0)
    }

    return xs
}