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) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		sc.nextLine();
		String s = sc.nextLine();

		List<Integer> a = new ArrayList<>();
		int current = 1;
		int count = 0;
		for (int i = 0; i < n; i++) {
			if (s.charAt(i) == '0' + current) count++;
			else {
				current = 1 - current;
				a.add(count);
				count = 1;
			}
		}

		a.add(count);
		if (a.size() % 2 == 0) a.add(0);

		int[] sum = new int[a.size() + 1];
		for (int i = 1; i < a.size() + 1; i++) {
			sum[i] = sum[i - 1] + a.get(i - 1);
		}

		int answer = 0;
		for (int i = 0; i < a.size(); i += 2) {
			int l = i;
			int r = Math.min(l + 2 * k + 1, a.size());
			answer = Math.max(answer, sum[r] - sum[l]);
		}
		System.out.println(answer);
	}
}
