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 (1.3.71) (judge verdict: Wrong Answer)

import kotlin.math.max

//

fun main() {
    abc124d()
}

fun abc124d() {
    val (n, k) = readLine()!!.split(" ").map { it.toInt() }
    val s = readLine()!!
    val table = List(2) { IntArray(n + 1) { 0 } }
    val startZero = s[0] == '0'
    var z = 0
    var o = 0
    var performZero = startZero
    for (c in s) {
        if (c == '0') {
            if (!performZero) {
                // 連続する1が途切れたので1のindexをすすめる
                performZero = true
                o++
            }
            table[0][z]++
        } else {
            if (performZero) {
                // 連続する0が途切れたので、ゼロのindexをすすめる
                performZero = false
                z++
            }
            table[1][o]++
        }
    }
    var count = 0
    for (i in 0 until k) {
        count += table[0][i] + table[1][i]
        if (table[0][i] == 0 && table[1][i] == 0) break
    }
    if (!startZero) {
        count += table[1][k]
    }
    var maxCount = count
    for (i in k until n - k) {
        if (table[0][i] == 0 && table[1][i] == 0) break
        count -= table[0][i - k]
        count -= if (startZero && i - k - 1 > 0) table[1][i - k - 1] else table[1][i - k]
        count += table[0][i]
        count += if (startZero) table[1][i] else table[1][i + 1]
        maxCount = max(count, maxCount)
    }
    println(maxCount)
}