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: Time Limit Exceeded)

#include <stdio.h>

int main(){
   char s[100001],c;
   int i,n,w[100002],j,k,l,ii,kk,ml;
    
   scanf("%d %d",&n,&k);
   scanf("%s",s);
   c=s[0];j=0;l=1;
   for (i=1;i<n;i++)
      if (s[i]!=c) {
         if (c=='0') w[j++]=-l;
         else        w[j++]=l;
         c=s[i];
         l=1;
      } else l++;
   w[j++]=l;
   ml=0;
   for (i=0;i<j;i++) {
      kk=k;l=0;
      for (ii=i;ii<j && kk>0;ii++)
         if (w[ii]<0) {
            kk--;l+=-w[ii];
         } else
            l+=w[ii];
      if (ii<j && w[ii]>0) l+=w[ii];
      if (ml<l) ml=l;
   }
   printf("%d\n",ml);
   return 0;
}