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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <string.h>

int main()
{
  int n,k,a[100100],b=0;
  scanf("%d %d",&n,&k);
  char s[100100];
  scanf("%s",s);
  int i=0;
  while(i<n)
    {
      while(i<n&&s[i] == '1')
	{
	  a[b]++;
	  i++;
	}

      b++;
      
   while(i<n&&s[i] == '0')
	{
	  a[b]++;
	  i++;
	}
   b++;
    };

  k = 2*k+1;
  int now,max=0;
  for(i=0;i<k;i++)
    max += a[i];

  if(max == n)
    printf("%d\n",max);

  else
    {

  now = max;

  for(i=k;i<b;i+=2)
    {
    now += a[i] + a[i+1] - a[i-k] - a[i-k+1];
    if(now > max)
      max=now;
    }

  printf("%d\n",max);
    }
  return 0;
}