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 {
 
    private static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {

        int n = Integer.parseInt(sc.next());
        int k = Integer.parseInt(sc.next());
        String s=sc.next();
        char[] si  = s.toCharArray();
        char s0='0';
        char s1='1';
        ArrayList<Integer> numS =new ArrayList<Integer>();
        int now = 1;
        int count=0;
        //101111011->11412へ
        for(char curr:si){
            if(Character.getNumericValue(curr) == now){
                count++;
            }else{
                numS.add(count);
                now=1-now;
                count=1;
            }
        }
        if(count!=0)numS.add(count);
        if(numS.size()%2==0)numS.add(0);
        //System.out.println(numS);
        int add = 2 * k + 1;
        int ans = 0;
        //11412
        int left =0;
        int right =0;
        int tmp =0;
        for(int nextleft = 0; nextleft < numS.size(); nextleft += 2)
        {
            int nextright = Math.min( nextleft + add, numS.size() );

            while(nextleft > left)
            {
                tmp -=numS.get(left);
                left++;
            }
            while(nextright > right)
            {
                tmp +=numS.get(right);
                right++;
            }
            ans = Math.max(tmp,ans);
        }

        System.out.println(ans);
    }
}