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 {

    static Scanner sc = new Scanner(System.in);


    public static void main(String[] args) {
        int n = sc.nextInt();
        int k = sc.nextInt();
        String s = sc.next();
        List<Integer> list = new ArrayList<>();
        if (s.charAt(0) == '0') {
            list.add(0);
        }
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            if (i == 0 || s.charAt(i) == s.charAt(i - 1)) {
                cnt++;
            } else {
                list.add(cnt);
                cnt = 1;
            }
        }
        list.add(cnt);
        if (list.size() % 2 == 0) {
            list.add(0);
        }
        int ans = 0;
        for (int i = 0; i < list.size(); i += 2) {
            int left = i;
            int right = Math.min(i + k * 2 + 1, list.size());
            int num = 0;
            for (int j = left; j < right; j++) {
                num += list.get(j);
            }
            if (ans < num) {
                ans = num;
            }
        }

        System.out.println(ans);
    }
}
