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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        String[] tmp = br.readLine().split(" ");
        int n = Integer.parseInt(tmp[0]);
        int k = Integer.parseInt(tmp[1]);
        char[] s = br.readLine().toCharArray();

        List<Integer> list = new ArrayList<>();
        boolean b = true;
        int i;
        int count = 0;
        for(i = 0; i < n; i++) {
            if(b != (s[i] == '1')) {
                list.add(count);
                count = 1;
                b = !b;
            }else {
                count++;
            }
        }
        list.add(count);
        if(list.size() % 2 == 0) list.add(0);


        int ans = 0;
        int j = 0;
        do {
            int y = 0;
            for(int x = 2 * j; x < list.size() && x <= 2 * j + 2 * k; x++) {
                y += list.get(x);
            }
            j++;
            ans = Math.max(ans, y);
        }while(j < list.size() / 2);

        System.out.println(ans);
    }
}