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

#define NN 100010

int main(void)
{
  int i, k, id, N, K, sum;
  scanf("%d %d",&N, &K);
  int s[NN];
  for(i=0;i<N;i++){
    scanf("%01d",&s[i]);
  }
  /**/
  int len[N+2];
  int old = 1;
  int d = 0;  
  for(i=0;i<N+2;i++) len[i] = 0;
  for(i=0;i<N;i++){
    if(old !=s[i] ) d ++;
    len[d] ++; 
    old = s[i];
  }
  if(s[N-1] == 0){
    d++;
    len[d]=0;
  }
  int L = d+1;
  int max = 0;
  // if K==0
  if(K==0){
    for(i=0;i<L;i+=2){
      if (len[i] > max ){
	max = len[i];
	id = i;
      }
    }
    printf("%d", max); 
    return (0);
  }
  //-----
    
   // --max
   int sumL, sumR;
   int idL, idR;
     
   sum = 0;
   max = 0;
   for(i=1;i<L;i+=2){
     sum = len[i-1] + len[i] + len[i+1] ;
     if (sum > max ){
       max = sum;
       id = i;
     }
   }
   // here anser of K=1 max
   /*
   if(id==1){
     idL = 1;
     idR = 1 + 2*(K-1);
   }
   */
   idL = id;
   idR = id;
   
   if(K> (L/2) ) K = L/2;
   
   if(K>=2){
     for(k=1;k<K;k++){
       
       if( (idL>=3)&&(idR<=(L-4))  ){       
	 sumL = len[idL-3] + len[idL-2] ;
	 sumR = len[idR+2] + len[idR+3] ;
	 
	 if( sumL < sumR  ){
	   idR += 2;
	 }else{
	   idL -= 2;
	 }
       }else if( (idL==1)&&(idR<=(L-4)) ){	   
	 idR+=2;
       }else if( (idL>=3)&&(idR==(L-2)) ){
	 idL-=2;
       }else if( (idL==1)&&(idR==(L-2)) ){
	 break;
       }else{

	 break;
       }
     }
   }
   //   printf(" Left %d Right %d\n", idL, idR); 
      
   max=0;
   for(i=idL-1;i<=idR+1;i++) max += len[i];
   printf("%d", max); 
   return (0);
}