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: Accepted)

#include<stdio.h>

int main(){
  int N, K;

  scanf("%d %d", &N, &K);

  int i;

  char S[N+10];

  scanf("%s", S);

  int X[N];

  for(i = 0; i < N; i++){
    X[i] = S[i] - 48;
  }

  int count[100005] = {};
  int c = 0, j = 0;

  int ins = X[0];

  for(i = 0; i < N; i++){
    if(ins != X[i]){
      ins = X[i];
      j++;
    }
    count[j]++;
  }
  
  int max = 0;

  i = 0; c = 0; j = 0; ins = 0;
  
  while(count[i] != 0){
    if(j < K){;
      if(i % 2 == X[0]){
        j++;
      }
      ins += count[i];
      if(max < ins) max = ins;
      i++;
    }
    else{
      if(i % 2 == X[0]){
        if(X[0] == 1){
          ins -= count[2 * c] + count[2 * c + 1];
        }
        else{
          if(c != 0) ins -= count[2 * c - 1];
          ins -= count[2 * c];
        }
        c++;
      }
      ins += count[i];
      i++;
      if(max < ins) max = ins;
    }
  }

  printf("%d\n", max);

  return 0;
}