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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		try (Scanner s = new Scanner(System.in)) {
			final int N = s.nextInt();
			final int K = s.nextInt();
			final String S = s.next();
			List<Integer> list = new ArrayList<>();
			char c = '1';
			int n = 0;
			for (int i = 0; i < N; i++) {
				if (c != S.charAt(i)) {
					list.add(n);
					c = S.charAt(i);
					n = 1;
				} else {
					n++;
				}
			}
			list.add(n);

			int ret = 0;
			for (int i = 0; i < list.size(); i += 2) {
				int END = i + 2 * K;
				int sum = 0;
				for (int j = i; j < list.size() && j <= END; j++) {
					sum += list.get(j);
				}
				ret = Math.max(ret, sum);
			}

			System.out.println(ret);
		}
	}
}
