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: Accepted)

#include<stdio.h>
int MAX(int a,int b){return a<b?b:a;}
int main(){
  int n,d[100010],s[100010],r=0,c,a,m,i,ss[100010]={};
  
  scanf("%d %d",&n,&m);
  for(i=0;i<n;i++)scanf("%1d",&d[i]);
  a=1;
  c=0;
  for(i=0;i<n;i++){
    if(d[i]-a){
      s[r++]=c;
      c=1;
    }
    else c++;
    a=d[i];
  }
  s[r++]=c;
  if(d[n-1]==0)s[r++]=0;
  //for(i=0;i<r;i++)printf("%d",s[i]);printf("\n");
  for(i=0;i<r;i++)ss[i+1]+=s[i]+ss[i];
  //for(i=0;i<r;i++)printf("%d ",ss[i]);printf("\n");
  for(i=a=0;i+m+m<=r;i+=2)a=MAX(ss[i+m*2+1]-ss[i],a);
  //printf("%d\n",r);
  if(m+m+1>r)a=n;
  printf("%d\n",a);
  return 0;
}