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()
{
	int N, K;
	char S[100001];
	scanf("%d %d", &N, &K);
	scanf("%s", S);
	
	int i, j, k, l[50001] = {}, r[50001] = {};
	for (i = 0, k = 0; i < N; ) {
		for (; S[i] == '0'; i++);
		l[++k] = i;
		for (; S[i] == '1'; i++);
		r[k] = i;
	}
	
	if (k <= K) {
		printf("%d\n", N);
		fflush(stdout);
		return 0;
	}

	int max = 0;
	for (i = 0; i <= k - K; i++) if (r[i+K] - l[i] > max) max = r[i+K] - l[i];
	printf("%d\n", max);
	fflush(stdout);
	return 0;
}