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>
int max2(int a, int b){return a<b?b:a;}
int n, m, i;
int k;
char s[100001];
int D[100002], *d, e;
int now, max;
int main(void){
  scanf("%d%d%s", &n, &k, s);
  if(n==1){puts("1"); return 0;}  
  e=1; m=1;
  for(i=1; i<n-1; i++){
    if(s[i]==s[i+1]){
      e++;
    }else{
      D[m]=e+1;
      e=0; m++;
    }
  }
  D[m]=e+1;
  e=0; m++;
  
  if(s[0]=='0'){d=D; m++;}else{d=D+1;}
  
  k = (k<<1)+1;
  
  if(k>=m){
    for(i=0; i<m; i++){
      now+=d[i];
    }
    printf("%d\n", now);
  }else{
    for(i=0; i<k; i++){
      now+=d[i];
    }
    max = now;
    for(i=k; i<m; i+=2){
      now+=d[i];
      now+=d[i+1];
      now-=d[i-k];
      now-=d[i-k+1];
      max = max2(max, now);
    }
    printf("%d\n", max);
  }
}