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.InputStreamReader;
import java.math.BigInteger;
import java.util.Scanner;

public class Main {

	private static final boolean TEST = false;

	public static void main(String[] args) {
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		while (true) {
			int n = in.nextInt();
			int k = in.nextInt();
			String s = in.next();
			System.out.println(solve(n, k, s));
			if (!TEST)
				break;
		}
		in.close();
	}

	private static int solve(int n, int k, String s) {
		int[] sv = new int[s.length()];
		int c = 1;
		int si = 0;
		for (int i = 1; i < s.length(); i++) {
			if (s.charAt(i) != s.charAt(i - 1))
				sv[si++] = c;
			c++;
		}
		sv[si] = c;

		int ans = 0;
		if (s.charAt(0) == '0') {
			ans = sv[Math.min(k * 2 - 1, si)];
			for (int i = 0; i <= si; i += 2)
				ans = Math.max(ans, sv[Math.min(i + 2 * k + 1, si)] - sv[i]);
		} else {
			ans = sv[Math.min(k * 2, si)];
			for (int i = 1; i <= si; i += 2)
				ans = Math.max(ans, sv[Math.min(i + 2 * k + 1, si)] - sv[i]);
		}
		return ans;
	}
}
