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.Arrays;
import java.util.Scanner;

public class Main {
    static Scanner scanner = new Scanner(System.in);

    public static void main(String[]$) {
        int n = scanner.nextInt(), k = scanner.nextInt();
        String s = scanner.next();
        int[] a = new int[s.split("1+").length + s.split("0+").length + (s.charAt(0) == '0' ? 1 : 0)];
        if (a.length == 1 || s.matches("0+")) {
            System.out.println(s.length());
            return;
        }
        Arrays.fill(a, 1);
        a[a.length - 1] = 0;
        int i = -1;
        if (s.charAt(0) == '0') {
            a[0] = 0;
            i++;
        }
        char last = ' ';
        for (char ch : s.toCharArray()) {
            if (ch == last) {
                a[i]++;
            } else {
                i++;
            }
            last = ch;
        }

        int[] sum = new int[a.length];
        for (int j = 0; j <= Math.min(k * 2, a.length - 1); j++) {
            sum[0] += a[j];
        }
        for (int j = 2; j < a.length - k * 2; j += 2) {
            sum[j / 2] = sum[j / 2 - 1] - a[j - 1] - a[j - 2] + a[j + k * 2 - 1] + a[j + k * 2];
        }

        Arrays.stream(sum).max().ifPresent(System.out::println);

    }
}