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.

Examples

Input

5 1
00010


Output

4


Input

14 2
11101010110011


Output

8


Input

1 1
1


Output

1
#include <iostream>
#include <algorithm>

using namespace std;
int main()
{
	int N, K;
	string s;	
	cin >> N >> K;
	cin >> s;
	int left = 0;
	int longest = 1;
	int i = 1;
	int cnt = (s[0] == '0') ? 1 : 0;
	while(i < N && N - left > longest)
	{
		for(; i < N; i += 1)
		{
			if(s[i] == '0' && s[i-1] == '1')
			{
				cnt += 1;
				if(cnt > K)
				{
					break;
				}
			}
		}
		longest = max(longest, i - left);

		left += 1;
		while(left < N && !(s[left - 1] == '0' && s[left] == '1'))
		{
			left += 1;
		}
		cnt -= 2;
	}

	cout << longest << endl;
	return 0;
}
