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 sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        String s = sc.next();

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

        char prev = s.charAt(0);
        if (prev == '0') {
            nums.add(0);
        }

        int count = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == prev) {
                count++;
            } else {
                nums.add(count);
                prev = s.charAt(i);
                count = 1;
            }
        }
        if (count != 0) {
            nums.add(count);
        }
        if (prev == '0') {
            nums.add(0);
        }

        int c = 0;
        int[] sum = new int[nums.size() + 1];
        int i = 1;
        for (Integer num : nums) {
            sum[i] = sum[i - 1] + num;
            i++;
        }

        int result = 0;
        for (int l = 0; l < sum.length; l += 2) {
            int r = l + 2 * k + 1;
            if (r >= sum.length) r = sum.length - 1;
            result = Math.max(result, sum[r] - sum[l]);
        }

        System.out.println(result);
    }
}
