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
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;

public class Main {

	public static void main(String args[]) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String str[] = br.readLine().split(" ");

		int n = Integer.parseInt(str[0]);
		int k = Integer.parseInt(str[1]);

		str = br.readLine().split("");
		ArrayList<Integer> nums = new ArrayList<>();

		if (str[0].equals("0"))
			nums.add(0);
		for (int i = 0; i < str.length;) {
			int j = i;
			while (j < str.length && str[j].equals(str[i]))
				++j;
			nums.add(j - i);
			i = j;
		}
		if (str[n - 1].equals("0"))
			nums.add(0);

		int s[] = new int[nums.size() + 1];

		for (int i = 0; i < nums.size(); i++) {
			s[i + 1] = s[i] + nums.get(i);
		}

		int ans = 0;
		for (int left = 0; left < nums.size(); left += 2) {
			int right = left + k * 2 + 1;
			if (right >= s.length)
				right = s.length - 1;
			ans = Math.max(ans, s[right] - s[left]);
		}
		System.out.println(ans);
	}
}