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 scan=new Scanner(System.in);
		int n=scan.nextInt();
		int k=scan.nextInt();
		String S=scan.next();
		
		int head[]=new int[n+1];
		int count=1;
		for(int i=1;i<n;i++)
		{
			if(S.charAt(i-1)!=S.charAt(i))
			{
				head[count++]=i;
			}
		}
		head[count]=n;
		int max=1;
		for(int i=0;i<count;i++)
		{
			boolean zero=S.charAt(head[i])=='0';
			int last=Math.min(i+k*2+(zero?0:1),count);
			int sum=head[last]-head[i];
			max=Math.max(max, sum);
		}
		System.out.println(max);
	}
}
