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(String::toInt)
    val a = readLine()!!.toCharArray()
    val changePoints = mutableListOf(0)
    (1 until n).forEach { if (a[it - 1] != a[it]) changePoints.add(it) }
    changePoints.add(n)
    var max = 0
    var lp = 0
    while (true) {
        val rp = lp + 2 * k - if (a[changePoints[lp]] == '0') 1 else 0
        if (rp >= changePoints.size - 1) {
            max = Math.max(changePoints.last() - changePoints[lp], max)
            break
        }
        max = Math.max(changePoints[rp + 1] - changePoints[lp], max)
        lp++
    }
    println(max)
}