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>
#include <stdlib.h>


int main()
{
	int n, k, start = 0, end = 0;
	char s[100001];
	scanf("%d %d", &n, &k);
	fflush(stdin);
	gets(s);
	int max = 0;
	for (int i = 1; i <= k; i++)
	{
		for (int j = 0; j <n; j++)
		{
			int len = 0, right, left = j - 1, st = j;
			if (s[j] == '0')
			{
				while (s[j] == '0' && j < n)
				{
					len++;
					j++;
				}
				right = j;
				j--;
				while (s[right] == '1' && right < n)
				{
					len++;
					right++;
				}
				while (left >= 0 && s[left] == '1' && left < n)
				{
					len++;
					left--;
				}
				if (len > max)
				{
					max = len;
					start = st;
					end = j;
				}
			}
			else
			{
				while (s[j] == '1' && j < n)
				{
					j++;
					len++;
				}
				if (len > max)
					max = len;
				j--;
			}
			
		}
		for (int l = start; l <= end; l++)
			s[l] = '1';
	}
	printf("%d", max);
	return 0;
}