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 s  =new Scanner(System.in);
        int n = s.nextInt(), k = s.nextInt();
        String line = s.next();

        List<Integer> list = new ArrayList<>();

        {
            String line2 = line + "$";
            char c = line2.charAt(0);
            int cnt = 1;
            for (int i = 1; i < line2.length(); ++i) {
                if (c != line2.charAt(i)) {
                    list.add(cnt);
                    cnt = 1;
                    c = line2.charAt(i);
                } else {
                    ++cnt;
                }
            }
        }

        if (line.charAt(0) == '0') {
            list.add(0, 0);

        }
        if (line.charAt(line.length() - 1) == '0') {
            list.add(0);
        }

        int max = 0, cnt = 0;
        for (int i = 0; i < list.size(); ++i) {
            if (i  < k * 2 + 1) {
                cnt += list.get(i);
            } else if (i % 2 != 0) {
                cnt -= list.get(i - k * 2 - 1);
                cnt -= list.get(i - k * 2);
                cnt += list.get(i);
                cnt += list.get(i + 1);
            }
            max = Math.max(max, cnt);
        }
        System.out.println(max);
    }
}
