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 PHP (7.4.4) (judge verdict: Accepted)

<?php
  list($n, $k) = explode(" ", trim(fgets(STDIN)));
  list($s) = explode(" ", trim(fgets(STDIN)));

  $count = [];
  $pa = 0;
  $count[$pa]=0;
  $cZero = 0;
  $zen = '1';
  for ($i=0; $i<$n; $i++) {
    if($zen == $s[$i]){
      $count[$pa]++;
    }else{
      $pa++;
      $zen = $s[$i];
      if($zen == '0') {
        $cZero++;
      }
      $count[$pa]=1;
    }
  }

  if($cZero <= $k){
    echo $n;
    exit;
  }

  $max = 0;
  for ($i=0; $i<$k*2+1; $i++) {
    $max += $count[$i];
  }

  $sum = $max;
  if($pa % 2 == 1) {
    $pa++;
    $count[$pa]=0;
  }
  for ($i=$k*2+2; $i<=$pa; $i=$i+2) {
//    echo $sum.PHP_EOL;
    $sum+=$count[$i];
    $sum+=$count[$i-1];
    $sum-=$count[$i-$k*2-1];
    $sum-=$count[$i-$k*2-2];
    $max = max($max, $sum);
  }

  echo $max;