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 sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        char[] s = sc.next().toCharArray();
        ArrayDeque<Integer> ad = new ArrayDeque<>();
        ad.add(0);
        boolean zeroStarted = false;
        int zeroSec = 0;
        int ans = 0;
        for (int i = 0; i < n; i++) {
            if (s[i] == '0') {
                if (!zeroStarted) {
                    zeroStarted = true;
                    if (zeroSec >= k) {
                        ans = Math.max(ans, i - ad.poll());
                        continue;
                    }
                }
            }else{
                if (zeroStarted) {
                    zeroStarted = false;
                    ad.add(i);
                    if(zeroSec < k) zeroSec++;
                }
            }
        }
        ans = Math.max(ans, n - ad.poll());
        System.out.println(ans);
        sc.close();

    }

}
