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>
void dump_arr (int arr[], int len) {
  for (int i = 0; i < len; i++) {
    printf("[%d] = %d\n", i, arr[i]);
  }
}

#define min(a, b) ((a) < (b)) ? (a) : (b)
#define max(a, b) ((a) > (b)) ? (a) : (b)

int main(int argc, const char *argv[])
{
  int n, k;
  char s[100001];
  int m[100002], mc = 0;
  int sum_m[100003];
  scanf("%d%d%s", &n, &k, s);

  char cur, last_cur;
  int count = 0;
  cur = s[0];
  last_cur = s[0];
  if (s[0] == '0') {
    m[mc++] = 0;
  }
  for (int i = 1; i <= n; i++) {
    cur = s[i];
    count++;
    if (cur != last_cur){
      m[mc++] = count;
      count = 0;
    }
    last_cur = cur;
  }
  // Array must be odd number like 1...1
  if (mc % 2 == 0)
    m[mc++] = 0;

  // 累積和
  sum_m[0] = 0;
  for (int i = 0; i < mc; i++) {
    sum_m[i+1] = sum_m[i] + m[i];
  }
  // dump_arr(sum_m, mc);

  int tmp_sum = 0, max_sum = 0;
  // 0, 2, 4 ... even number is zero
  for (int a = 0; a < mc; a += 2) {
    int right = a + k * 2 + 1;
    int left = a;
    if ( right > mc )
      right = mc;
    // printf("[%d] - [%d]\n", right, left);
    tmp_sum = sum_m[right] - sum_m[left];
    max_sum = max(max_sum, tmp_sum);
  }
  printf("%d\n", max_sum);
  // dump_arr(m, mc);
  return 0;
}