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.*;

class Main {
  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);

    int N = in.nextInt();
    int K = in.nextInt();
    char[] s = in.next().toCharArray();

    List<Integer> nums = new ArrayList<Integer>();

    char prev = s[0];
    int count = 1;

    for (int i = 1; i < N; i++) {
      if (prev == s[i]) {
        count++;
      } else {
        nums.add(count);
        count = 1;
        prev = s[i];
      }
    }
    nums.add(count);

    // Make nums start with '0' count
    if (s[0] == '1') {
      nums.add(0, 0);
    }
    // Make num ends with '0' count
    if (s[N - 1] == '1') {
      nums.add(0);
    }

    int numflip = 0;
    int ans = 0;
    int curr = 0;

    for (int i = 0; i < nums.size(); i++) {
      // We are looking at '0' count
      if (i % 2 == 0) {
        if (numflip < K) {
          numflip++;
        } else {
          int kickedOut0Index = i - K * 2;
          curr -= nums.get(kickedOut0Index);
          if (0 < kickedOut0Index) {
            curr -= nums.get(kickedOut0Index - 1);
          }
        }
        curr += nums.get(i);
      }
      // We are looking at '1' count
      else {
        curr += nums.get(i);
      }
      ans = Math.max(ans, curr);
    }

    System.out.println(ans);

  }
}