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: Time Limit Exceeded)

#include <stdio.h>

#define N 100000

int main() {
	static char s[N + 1];
	int n, k, k_, i, j, ans;

	scanf("%d%d%s", &n, &k, s);
	ans = 0;
	for (i = 0; i < n; ) {
		k_ = k;
		j = i + 1;
		while (j < n) {
			if (s[j] != s[j - 1])
				if (s[j] == '0' && k_ == 0)
					break;
				else if (s[j] == '1')
					k_--;
			j++;
		}
		if (ans < j - i)
			ans = j - i;
		for (j = i + 1; j < n && s[j] != '1'; j++)
			;
		i = j;
		if (ans == n) {
			printf("%d\n", ans);
			return 0;
		}
	}
	printf("%d\n", ans);
	return 0;
}