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 len = sc.nextInt();
        int k = sc.nextInt();
        String str = sc.next();
        int leftIdx = 0;
        int cnt = 0;
        int maxLen = 0;
        for(int i = 0; i < len; i++) {
          char ch = str.charAt(i);
          if(ch == '0') {
            cnt++;
            while(i < len - 1 && str.charAt(i+1) == '0') {
              i++;
            }
          }
          while(cnt > k) {
            if(str.charAt(leftIdx) == '0') {
              cnt--;
              while(leftIdx < i && str.charAt(leftIdx) == '0') {
                leftIdx++;
              }
            } else {
              leftIdx++;
            }
          }
          maxLen = Math.max(maxLen, i - leftIdx + 1);
        }
        System.out.println(maxLen);
    }
}