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;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        String s = sc.next();
        int cnt = 1;
        char c = s.charAt(0);
        List<Integer> list = new ArrayList<>();
        if (c == '0') {
            list.add(0);
        }
        for (int i = 1; i < n; i++) {
            char tmp = s.charAt(i);
            if (c == tmp) {
                cnt++;
                continue;
            }
            list.add(cnt);
            cnt = 1;
            c = tmp;
        }
        list.add(cnt);
        if (s.charAt(n - 1) == '0') {
            list.add(0);
        }
        int size = list.size();
        long ar[] = new long[size];
        for (int i = 0; i < size; i++) {
            ar[i] = list.get(i);
        }
        for (int i = 1; i < size; i++) {
            ar[i] += ar[i - 1];
        }
        long ans = ar[Math.min(size - 1, k * 2)];
        for (int r = k * 2 + 2; r < size; r += 2) {
            long tmp = ar[r] - ar[r - (k * 2 + 1)];
            ans = Math.max(ans, tmp);
        }
        System.out.println(ans);
        sc.close();

    }
}
