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: Runtime Error)

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

char S[100000];
int s[50000];

void Do2(int N, int n, int K){
  int i, M, m;
  M = 0;
  m = 0;
  for(i = 0;i <= 2 * K - 1;i++){
    if(i >= n) break;
    m += s[i];
  }
  M = m;
  if(M != N){
    for(i = 1;i < n;i += 2){
      if(i + 2 * K - 1 >= n) break;
      m -= s[i-1];
      if(i >= 2) m -= s[i-2];
      m += s[i+2*K-1];
      if(i + 2 * K < n) m += s[i+2*K];
      if(M < m) M = m;
    }
  }
  printf("%d\n", M);
}

void Do(int N, int K){
  int i, c, C;
  c = C = 0;
  for(i = 0;i < N;i++){
    if(S[i] == '0' + (C % 2)) c++;
    else{
      s[C] = c;
      C++;
      c = 1;
    }
  }
  s[C] = c;
  C++;
  Do2(N, C, K);
}

int main(void){
  int N, K;
  scanf("%d %d", &N, &K);
  scanf("%s", S);
  Do(N, K);
  return 0;
}