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 = Integer.parseInt(sc.next());
		int K = Integer.parseInt(sc.next());
		String S = sc.next();
		
		int cnt = 0;
		int ans = 0;
		int c = 0;
		ArrayList<Integer> ary = new ArrayList<Integer>();
		ary.add(0);
		cnt++;
		
		for(int i = 0; i < N - 1; i++) {
			if(!S.substring(i, i + 1).equals(S.substring(i + 1, i + 2))) {
				if(S.substring(i, i + 1).equals("1")) {
					if(cnt > K) {
						cnt--;
						ans = Math.max(ans, i + 1 - ary.get(c));
						c++;
					}
				} else {
					ary.add(i + 1);
					cnt++;
				}
			}
		}
		ans = Math.max(ans, N - ary.get(c));
		System.out.println(ans);
	}
}
