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.util.*;

public class Main{

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), k = sc.nextInt();
		String s = sc.next();
		sc.close();
		int ans = 0;
		char prev = '1';
		int cnt[] = new int[n + 2 * k + 1];
		int j = 0;
		for(int i = 0; i < n; ++i) {
			if(s.charAt(i) == prev)cnt[j]++;
			else {
				prev = s.charAt(i);
				cnt[++j]++;
			}
		}
		int tmp = 0;
		for(int i = 0; i < 2 * k; ++i)
			tmp += cnt[i];
		for(int i = 2 * k; i < cnt.length; ++i) {
			tmp += cnt[i];
			if(i % 2 == 0)ans = Math.max(ans, tmp);
			tmp -= cnt[i - 2 * k];
		}
		System.out.println(ans);
		
	}

}
