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 scanner = new Scanner(System.in);
        String[] nk = scanner.nextLine().split(" ");
        int n = Integer.parseInt(nk[0]);
        int k = Integer.parseInt(nk[1]);
        String s = scanner.nextLine();

        List<Integer> a = new ArrayList<>();
        a.add(0);

        char current = s.charAt(0);
        for (int i = 1; i < n; i++) {

            if (s.charAt(i) != current) {
                a.add(i);
                current = s.charAt(i);
            }
        }
        a.add(s.length());


        int max = 0;


        for (int i = 0; i < a.size() - 1; i++) {

            int j = s.charAt(a.get(i)) == '0' ? i + 2 * k : i + 2 * k + 1;
            j = Math.min(j, a.size() - 1);

            int x = a.get(j) - a.get(i);

            if (x > max) {
                max = x;
            }
        }

        System.out.println(max);
    }
}