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(void) {

  long n,k;
	scanf("%ld %ld", &n, &k);
	char s[n+1];
	scanf("%s", s);
	long zero_list[n][2];
	long cur = 0;
	for (long i = 0; i < n; i++) {
		if (i == 0 && s[i] == '0') {
			zero_list[cur][0] = i;
		}
		if (i != 0 && s[i] == '0' && s[i-1] == '1') {
			zero_list[cur][0] = i;
		}
		if (i == n-1 && s[i] == '0') {
			zero_list[cur][1] = i;
			cur++;
		}
		if (i != n-1 && s[i] == '0' && s[i+1] == '1') {
			zero_list[cur][1] = i;
			cur++;
		}
	}
	if (k >= cur) {
		printf("%ld\n", n);
		return 0;
	}
	long max = zero_list[k][0];
	long length;
	for (long i = k+1; i < cur; i++) {
		length = zero_list[i][0]-zero_list[i-k-1][1]-1;
		if (length > max) {
			max = length;
		}
	}
	length = n-zero_list[cur-k-1][1]-1;
	if (length > max) {
		max = length;
	}
	printf("%ld\n", max);

  return 0;
}