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 static java.lang.System.*;
import java.util.*;

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

        int n = sc.nextInt();
		int k = sc.nextInt();
		char[] s = sc.next().toCharArray();
		
		ArrayList<Integer> list = new ArrayList<Integer>();
		
		char prev = '1';
		list.add(0);
		for(int i=0; i<n; i++) {
			if(s[i] == prev) {
				list.set(list.size()-1, list.get(list.size()-1)+1);
			}
			else {
				prev = s[i];
				list.add(1);
			}
		}
		for(int i=0; i<k*2; i++) list.add(0);		
			
		int ans = 0;
		int sum = 0;
		for(int i=0; i<k*2; i++) 
			sum += list.get(i);
		
		for(int i=k*2; i<list.size(); i++) {
			sum += list.get(i);			
			if(i%2==0) ans = Math.max(ans, sum);
			sum -= list.get(i-k*2);		
		}
		
        out.println(ans);
    }
}