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>

#define MAX 100002

int main()
{
  int n, k, i, j, ans[MAX], count = 0, pos = 0, sum, x, y, z;
  char s[MAX], flag;

  scanf("%d", &n);
  scanf("%d", &k);
  scanf("%s", s);

  memset(ans, 0, sizeof(ans));

  flag = s[0];

  for (i = 0; i < n; i++) {
    if (s[i] == flag) {
      count++;
    } else {
      ans[pos++] = count;
      count = 1;
      flag = s[i];
    }
  }

  ans[pos] = count;

  if (s[0] == '1' && s[n - 1] == '1') {
    x = 0;
    for (i = 0; i <= pos - k * 2; i += 2) {
      sum = 0;
      for (j = i; j < i + k * 2 + 1; j++) {
	sum += ans[j];
      }
      if (x < sum) {
	x = sum;
      }
    }
  } else if (s[0] == '0' && s[n - 1] == '1') {
    x = 0;
    for (i = 0; i < k * 2; i++) {
      x += ans[i];
    }
    for (i = 1; i <= pos - k * 2; i += 2) {
      sum = 0;
      for (j = i; j < i + k * 2 + 1; j++) {
	sum += ans[j];
      }
      if (x < sum) {
	x = sum;
      }
    }
  } else if (s[0] == '1' && s[n - 1] == '0') {
    x = 0;
    for (i = pos; i > pos - k * 2; i--) {
      x += ans[i];
    }
    for (i = 0; i <= pos - k * 2; i += 2) {
      sum = 0;
      for (j = i; j < i + k * 2 + 1; j++) {
	sum += ans[j];
      }
      if (x < sum) {
	x = sum;
      }
    }
  } else {
    y = 0;
    z = 0;
    for (i = 0; i < k * 2; i++) {
      y += ans[i];
    }
    for (i = pos; i > pos - k * 2; i--) {
      z += ans[i];
    }
    x = (y > z ? y : z); 
    for (i = 1; i <= pos - k * 2; i += 2) {
      sum = 0;
      for (j = i; j < i + k * 2 + 1; j++) {
	sum += ans[j];
      }
      if (x < sum) {
	x = sum;
      }
    }
  }

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

  return 0;
}