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[] $){
		Scanner s=new Scanner(System.in);

		int n=s.nextInt(),k=s.nextInt();
		char[]t=s.next().toCharArray();

		ArrayList<Integer>l=new ArrayList<>();
		if(t[0]=='0')
			l.add(0);
		{
			int c=1;
			for(int i=0;i<n;++i) {
				if(i==n-1||t[i+1]!=t[i]) {
					l.add(c);
					c=1;
				}else
					++c;
			}
		}
		if(t[n-1]=='0')
			l.add(0);

		if(k*2+1>=l.size()) {
			System.out.println(t.length);
			return;
		}

		int sum=0,r=0;
		for(int i=0;i<=(k-1)*2;++i)
			sum+=l.get(i);

		for(int i=k*2;i<l.size();i+=2) {
			sum+=l.get(i);
			sum+=l.get(i-1);

			System.err.println(sum);
			r=Math.max(r,sum);

			sum-=l.get(i-k*2);
			sum-=l.get(i-k*2+1);
		}

		System.out.println(r);
	}
}