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: Wrong Answer)

fun main(args :Array<String>) {
    val (n, k) = readIntegerList()
    val s = readLine()!!
    val tmp = arrayListOf<Int>()
    var pre = s[0]
    var count = 1
    for (c in s.substring(1)) {
        if (pre == c) {
            count++
        } else {
            tmp.add(count)
            count = 1
        }

        pre = c;
    }
    tmp.add(count)

    if (s[0] == '0') {
        tmp.add(0, 0)
    }
    // println(tmp)
    var max = tmp
        .subList(0, Math.min(2 * k + 1, tmp.size))
        .sum()
    var a = max
    // println(a)
    for (i in 2 until tmp.size - (2 * k) step 2) {
        a = a - tmp[i - 1] - tmp[i - 2] + tmp[i + 2 * k - 1] + tmp[i + 2 * k]
        // println("${i - 2} ${i + 2 * k} ${a}")
        max = Math.max(max, a)
    }
    println(max)
}

fun readInteger() = readLine()!!.toInt()
fun readStringList() = readLine()!!.split(" ")
fun readIntegerList() = readStringList().map(String::toInt)