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

fscanf(STDIN, "%d %d", $n, $k);
fscanf(STDIN, "%s", $s);
$n = strlen($s);

// 何個まで選べるか
if ($s[0] === '1') {
    $K = 2 * $k + 1;
} else {
    $K = 2 * $k;
}

// 塊に分割
$blob = [];
$cnt = 0;
$prev = $s[0];
for ($i = 0; $i < $n; ++$i) {
    if ($s[$i] !== $prev) {
        $blob[] = $cnt;
        $cnt = 1;
    } else {
        ++$cnt;
    }

    $prev = $s[$i];
}
$blob[] = $cnt;

if (count($blob) < $K) {
    echo $n, PHP_EOL;
    return;
}

$bn = count($blob);
$sum = 0;
$max = 0;
$r = 0;
for ($l = 0; $l < $bn; ++$l) {
    while ($r < $bn && ($r - $l < $K)) {
        $sum += $blob[$r];
        ++$r;
    }
/* echo "l: $l, r: $r, sum: $sum", PHP_EOL; */
    if ($r > $bn) break;

    $max = max($max, $sum);

    $sum -= $blob[$l];
}

echo $max;