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();
    String s = sc.next();
    ArrayList<Integer> list = new ArrayList<Integer>();
    list.add(0);
    for(int i = 1; i < s.length(); i++) {
      if(s.charAt(i) != s.charAt(i - 1)) list.add(i);
    }
    int p = list.size();
    for(int i = 0; i < 2 * k + 1; i++) {
      list.add(n);
    }
    Collections.sort(list);
    int ans = 0;
    for(int i = 0; i < p; i++) {
      if(s.charAt(list.get(i)) == '0') {
        ans = Math.max(ans, list.get(i + 2 * k) - list.get(i));
      } else {
        ans = Math.max(ans, list.get(i + 2 * k + 1) - list.get(i));
      }
    }
    System.out.println(ans);
  }
}