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.*;
import java.util.stream.Collectors;

public class Main implements Runnable {

    private static int MOD = 1_000_000_007;

    public static void main(String[] args) {
        // Run with 16MB stack
        Thread thread = new Thread(null, new Main(), "", 16 * 1024 * 1024);
        thread.start();
    }

    @Override
    public void run() {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }

    // int[] a = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
    static void solve(Scanner scanner) {
        int[] a = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();
        int N = a[0], K = a[1];
        String S = scanner.nextLine();
        if (!S.contains("0") || !S.contains("1")) {
            System.out.println(N);
            return;
        }

        S = "1".concat(S).concat("1");
        List<Integer> starts = new ArrayList<>();
        List<Integer> ends = new ArrayList<>();
        boolean one = false;
        for (int i = 0; i < S.length(); i++) {
            if (S.charAt(i) == '1') {
                if (!one) {
                    one = true;
                    starts.add(i);
                }
            } else {
                if (one) {
                    ends.add(i - 1);
                    one = false;
                }
            }
        }
        if (one) {
            ends.add(S.length() - 1);
        }

        int max = 0;
        for (int i = 0; i < starts.size(); i++) {
            int endIdx = Math.min(i + K, ends.size() - 1);
            int ans = ends.get(endIdx) - starts.get(i) + 1;
            if (i == 0) {
                ans--;
            }
            if (endIdx == ends.size() - 1) {
                ans--;
            }
            max = Math.max(max, ans);
        }

        System.out.println(max);
    }

}
