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>
#define abs(x) ((x) > 0 ? (x) : -(x))
#define max(x, y) ((x) > (y) ? (x) : (y))

int
main(int argc, char *argv[])
{
  int n, k, ans;
  scanf("%d%d", &n, &k);
  int a, b = -1; int x[100005] = {}; int c = 0, minus = 0;
  for(int i = 0; i < n; i++){
    scanf("%1d", &a);
    if(b - a != 0) c++;
    if(a == 1) x[c]++;
    if(a == 0){
      x[c]--;
    }
    b = a;
  }
  x[0] = 0;
  x[++c] = 0;
  int sum[100005] = {};
  sum[0] = 0;
  for(int i = 1; i <= c; i++){
    sum[i] = sum[i - 1] + abs(x[i]);
    if(x[i] < 0) minus++;
  }
 

  if(minus <= k) ans = n;
  else{
    int t; ans = 0;
    if(x[1] < 0) t = 1;
    if(x[2] < 0) t = 2;
    while(t + 2 * (k - 1) < c){
      ans = max(ans, sum[t + 2 * (k - 1) + 1] - sum[max(t - 2, 0)]);
      t += 2;
    }
  }
  printf("%d\n", ans);
  return 0;
}