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: Accepted)

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

    var r = 0

    for (i in (0 until xs.size step 2)) {

        val l = if (i + 2 * k < xs.size) {
            i + 2 * k
        } else {
            xs.size - 1
        }

        val x = xs[l] - (if (i > 0) xs[i - 1] else 0)

        r = Math.max(r, x)
    }

    println(r)
}

private fun f(s: String): List<Int> {
    val xs = mutableListOf(0, 0)

    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(xs.last() + x)
            x = 0
        }
        x++
        c = it
    }

    xs.add(xs.last() + x)

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

    return xs
}