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 PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php
#Handstand

fscanf(STDIN, "%d %d", $N, $K);
$people = str_split(trim(fgets(STDIN)));

#最大K回までフリップできる。
#ceil(左から走査して最初に0を見つけたら1として以後数字が変わるごとにカウントした数 / 2)が必要（おそらく）。
#そのため、K回という数が全てを繋ぐために必要な数より少なければ、実際に計算する。
#その方法をどうやるか？

$req_k = req_k($people);
if ($K >= $req_k) {
    echo count($people) . "\n";
} 

echo $result . "\n";

#全員を逆立ちさせるのに必要な回数を求める
function req_k($people) {
    $count = 0;
    $number = 1;
    foreach ($people as $person) {
        if ($person != $number) {
            $count++;
            if ($number == 1) {
                $number = 0;
            } else {
                $number = 1;
            }
        }
    }
    return ceil($count / 2);
}


#連続した直立人数の最大値を求める
function calc_max_footstand($people) {
    $count = 0;
    $max_count = 0;
    foreach ($people as $person) {
        if ($person == 1) {
            $count = 0;
            continue;
        } else {
            $count++;
            if ($count > $max_count) $max_count = $count;
        }
    }
    return $max_count;
}

#連続した逆立ち人数の最大値を求める
function calc_max_handstand($people) {
    $count = 0;
    $max_count = 0;
    foreach ($people as $person) {
        if ($person == 0) {
            $count = 0;
            continue;
        } else {
            $count++;
            if ($count > $max_count) $max_count = $count;
        }
    }
    return $max_count;
}