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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

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

#define NN 100001

int len[NN];
//int iindex[NN];
int len1[NN][3];

void maxim(int *m, int *ii, int L)
{
  int max = 0;
  int sum , idi = 0;
  int i;
  for(i=1;i<L;i+=2){
    sum = len[i-1] + len[i] + len[i+1] ;
    if(max <sum ){
      max = sum;
      idi = i;
    }
  }
  *m = max;
  *ii = idi;
}

void oparate(int max, int idi, int L)
{
  int i;
  int old1, old2, old3;
  len[idi-1] = max;
  for(i=idi;i<L;i+=2){
    len[i] = len[i+2];
  }
}

int main(void)
{
  /*
  int N, K;
  scanf("%d %d ",&N, &K);
  char S[NN];
  scanf("%s",S);
  int i;
  int s[N];
  for(i=0;i<N;i++)  s[i] = S[i] -48;
  */
  int i, N, K;
  scanf("%d %d",&N, &K);
  int S;
  int s[N];
  for(i=0;i<N;i++){
    scanf("%01d",&s[i]);
  }
  //  for(i=0;i<N;i++) printf("%d ",s[i]);
  /**/
  int old = 1;
  int d=0;  
  for(i=0;i<N;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;
  /*
  printf("length = %d\n", L);  
  for(i=0;i<L;i++)   printf("%d ", len[i]); 
  printf("\n ");
*/
    
  int sum, e=0;
  int max = 0;
  int id;

  /* maxint */
   for(i=0;i<d;i+=2){
     if (len[i] > max ) max = len[i];
  }
    
  int k;
  //  for(i=0;i<length;i++){
    //    printf("%d(%d,%d,%d),  ",i, len1[i][0],  len1[i][1] , len1[i][2]); 
  //}
  for(k=0;k<K;k++){
    //    printf("\n ");
        //    printf("max %d  id%d \n", max, id);
    maxim(&max, &id, L);
    if(max >=N) break;
    L -=2;
    oparate(max, id, L);
    //  for(i=0;i<length;i++){
      //      printf("%d(%d,%d,%d),  ",i, len1[i][0],  len1[i][1] , len1[i][2]); 
    //}
    //    maxim(&max, &id, L);
  }
  //  maxim(&max, &id, length);

  printf("%d", max); 

  return (0);
}