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)

import java.util.*
import java.math.*
fun main(args: Array<String>) {

    val (N, K) = readLine()!!.split(" ").map(String::toInt)

    if (2 * K + 1 >= N) {
        print(N)
        return
    }
    var n = 0
    var isOder = false
    var ans = 0
    var tmp = IntArray(N){0}
    var from = 0
    var to = 0
    var tmpn = 0
    var k = K

    for (c in readLine()!!) {
        if (isOder) {
            if (c == '1') {
                isOder = false

                k--
                tmp[to++] = n
                tmpn += n

                n = 0
            }
            n++
        } else {
            if (c == '0') {
                isOder = true

                tmp[to++] = n
                tmpn += n
                if (k == 0) {
                    //println(tmp)
                    ans = Math.max(ans, tmpn)
                    tmpn -= tmp[from++] + tmp[from++]
                    k++
                }

                n = 0
            }
            n++
        }
    }
    if (n != 0) {
        //print(tmp)
        ans = Math.max(ans, tmpn+n)
    }


    //println(list)
    println(ans)
}