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 (5.6.30) (judge verdict: Wrong Answer)

<?php
/* fscanf(STDIN, "%d %d",$n,$k);
$s = strval(trim(fgets(STDIN)));
$array0 = array();
$array1 = array();
$sum0 = 0;
$sum1 = 0;
$total = 0;
$ans = 0;
$j = 0;
//0,1の連続で配列作成
for ($i=0;$i<=$n;$i++){
	if ($i == 0 && $s[$i] == 1){
		$sum1 ++;
	}elseif ($i == 0 && $s[$i] == 0){
		$sum0 ++;
	}else if ($s[$i] == $s[$i-1] && $s[$i] == 1){
		$sum1 ++;
	}else if ($s[$i] == $s[$i-1] && $s[$i] == 0){
		$sum0 ++;
	}else if ($s[$i] != $s[$i-1] && $s[$i] == 1){
		$array0 = $sum0;
		$sum0 = 0;
		$sum1 ++;
	}else if ($s[$i] != $s[$i-1] && $s[$i] == 0){
		$array1 = $sum1;
		$sum1 = 0;
		$sum0 ++;
	}
}
$arraySize = count ($array1);
for ($i=0;$i<$arraySize;$i++){
	if ($i == 0){
		for ($j=0;$j<=$k;$j++){
			$total += $sumArray[$j];
			$ans = $total;
		}
	}else {
		//左の尺を進める
		$total -= $sumArray[$i-1];
		//右の尺を進める
		$total += $sumArray[$k+1];
		
		if ($ans < $total){
			$ans = $total;
		}
	}
}
echo $ans; */

fscanf(STDIN, "%d %d",$n,$k);
$s = strval(trim(fgets(STDIN)));
$sumArray = array();
$sum = 0;
$total = 0;
$ans = 0;
$j = 0;
//0,1の連続で配列作成
for ($i=0;$i<=$n;$i++){
	if ($i == 0){
		$sum ++;
	}else if ($s[$i] == $s[$i-1]){
		$sum ++;
	}else{
		$sumArray[] += $sum;
		$sum = 1;
	}
}
$arraySize = count ($sumArray);
for ($i=0;$i<$arraySize;$i++){
	if ($i == 0){
		for ($j=0;$j<=$k;$j++){
			$total += $sumArray[$j];
			$ans = $total;
		}
	}else {
		//左の尺を進める
		$total -= $sumArray[$i-1];
		//右の尺を進める
		$total += $sumArray[$k+1];
		
		if ($ans < $total){
			$ans = $total;
		}
	}
}
echo $ans;
?>