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.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        final int N = in.nextInt();
        final int k = in.nextInt();
        final String s = in.next();

        int res = 0;
        int len = s.length();
        int left = 0;
        int countZeros = 0;

        for (int right = 0; right < len; ++right) {
            if (s.charAt(right) == '0') {
                countZeros++;
                for (int j = right; j < len && s.charAt(j) == '0'; j++) {
                    right = j;
                }
                while (countZeros > k) {
                    if (s.charAt(left) == '0') {
                        --countZeros;
                        for (int j = left; j < len && s.charAt(j) == '0'; j++) {
                            left = j;
                        }
                    }
                    left++;
                }

            }
            res = Math.max(res, right - left + 1);
        }
        System.out.println(res);
    }
}