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.Scanner;

public class Main {

    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        String s = sc.next();
        
        int l = 0;
        int r = 0;
        for(int i = 0; i < k; i++){
            r = s.indexOf("0", r);
            r = s.indexOf("10", r);
            if(r == -1){
                r = n-1;
                break;
            }
        }
        
        int max = 0;
        while(true){
            if(r-l+1 > max){
                max = r-l+1;
            }
            if(r == n-1) break;
            
            r = s.indexOf("0", r);
            r = s.indexOf("10", r);
            if(r == -1){
                r = n-1;
            }
            l = s.indexOf("01", l) +1;
        }
        
        System.out.println(max);
    }
}