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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Eric
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DHandstand solver = new DHandstand();
        solver.solve(1, in, out);
        out.close();
    }

    static class DHandstand {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int N = in.nextInt();
            int K = in.nextInt();
            String S = in.next();

            int res = shortSolve2(S, K);
            out.println(res);
        }

        private int shortSolve2(String s, int K) {
            int n = s.length();
            int i = 0;
            int j = 0;
            int res = 0;
            int k = 0;
            while (j < n) {
                while (j < n && s.charAt(j) == '1')
                    j++;
                res = Math.max(res, j - i);

                while (j < n && s.charAt(j) == '0')
                    j++;
                while (j < n && s.charAt(j) == '1')
                    j++;
                k++;
                if (k > K) {
                    while (i < n && s.charAt(i) == '1')
                        i++;
                    while (i < n && s.charAt(i) == '0')
                        i++;
                }
                res = Math.max(res, j - i);
            }
            return res;
        }

    }
}

