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>
#include <string.h>

struct count {
	int cnt;
	char moji;
};

int main() {
	int N, K, i, j, max = 0, sum = 0, k;
	char S[100001];
	struct count c[100000];

	memset(c, 0, sizeof(c));

	scanf("%d %d", &N, &K);
	scanf("%s", S);

	i = 0;
	j = 0;
	while (i < N) {
		c[j].moji = S[i];
		do {
			c[j].cnt++;
			i++;
		} while (i < N && c[j].moji == S[i]);
		j++;
	}
	
	k = K;
	for (i = 0; i < j && k > 0; i++) {
		if (c[i].moji == '0') k--;
		sum += c[i].cnt;
	}

	if (i == j) {
		printf("%d", sum);
		return 0;
	}
	
	sum += c[i].cnt;
	max = sum;
	if (i >= K * 2) sum -= c[i - K * 2].cnt;
	sum -= c[i - K * 2 + 1].cnt;

	for (i = i + 1; i < N; i++) {
		sum += c[i].cnt;
		if (++i < N) sum += c[i].cnt;
		if (sum > max) max = sum;
		sum -= c[i - K * 2].cnt;
		sum -= c[i - K * 2 + 1].cnt;
	}

	printf("%d", max);

	return 0;
}