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.Scanner;
import java.util.List;
import java.util.ArrayList;

public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        sc.nextLine();
        String S = sc.nextLine();
        sc.close();

        List<Integer> num = new ArrayList<Integer>();
        int obj = 1;
        int counter = 0;
        for (int i = 0; i < S.length(); i++) {
            if (S.charAt(i) == (char)('0' + obj)) {
                counter++;
            } else {
                num.add(counter);
                obj = 1 - obj;
                counter = 1;
            }
        }
        if (counter != 0) {
            num.add(counter);
        }
        if (num.size() % 2 == 0) {
            num.add(0);
        }

        int len = 2 * K + 1;
        int ans = 0;
        int temp;
        int sum[] = new int[num.size() + 1];
        sum[0] = 0;
        for (int i = 0; i < num.size(); i++) {
            sum[i + 1] = sum[i] + num.get(i);
        }
        for (int i = 0; i < num.size(); i += 2) {
            int left = i;
            int right = Math.min(i + len, sum.length - 1);
            temp = sum[right] - sum[left];
            ans = Math.max(ans, temp);
        }
        System.out.println(ans);
    }
}