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: Wrong Answer)

#include<stdio.h>
#include<math.h>
#include <stdlib.h>

int main()
{
  
  int i,j=0,N,K;
  scanf("%d %d",&N,&K);
  char S[N];
  int S_num[N];
  scanf("%s",S);
  
  S_num[0]=1;
  for(i=1;i<N;i++){
    S_num[i]=0;
  }
  
  for(i=1;i<N;i++){
    if(S[i]!=S[i-1]){
      j++;
    }
    S_num[j]++;
  }
  
  int S_cum[j+1];
  S_cum[0]=S_num[0];
  for(i=1;i<j+1;i++){
    S_cum[i]=S_cum[i-1] + S_num[i];
  }

  /*for(i=0;i<j+1;i++){
    printf("%d\n",S_cum[i]);
  }*/

  
  int length = j+1, max,temp;
  /*最初が0*/
  if(S[0]=='0'){
 
    if(length<=2*K){
      max=S_cum[length-1];
    } else {   
    /*最初から*/
      max=S_cum[2*K-1];
    /*途中から*/
    
      for(i=1;i < ((length+1)/2-K);i++){      
        temp=S_cum[2*i-1+2*K]-S_cum[2*i-2];
        if(temp>max){
        max=temp;
        }
      }
    /*最後*/
      if(length%2==0){
        temp=S_cum[length-1]-S_cum[length-2*K-2];
      }else{
        temp=S_cum[length-1]-S_cum[length-2*K-1];
      }
      if(temp>max){
        max=temp;
      }
    }
  }
    /*最初が1*/
  if(S[0]=='1'){
    
    if(length<=2*K-1){
      max=S_cum[length-1];      
    } else {   
    /*最初から*/
      max=S_cum[2*K];
    /*printf("%d\n",max);*/
    /*途中から*/
    
      for(i=1;i < (length/2-K);i++){      
        temp=S_cum[2*i+2*K]-S_cum[2*i-1];
        if(temp>max){
        max=temp;
        }
        /*printf("%d\n",temp);*/
      }
    /*最後*/
      if(length%2==0){
        temp=S_cum[length-1]-S_cum[length-2*K-1];
      }else{
        temp=S_cum[length-1]-S_cum[length-2*K-2];
      }
      if(temp>max){
        max=temp;
      }
      /*printf("%d\n",temp);*/
    }
  }
  
  printf("%d",max);
  return 0;

}