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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc=new Scanner(System.in);
		int N=sc.nextInt();
		int K=sc.nextInt();
		String s=sc.next();

		List<Integer> list=new ArrayList<>();

		int begin=0;
		char now='1';

		for(int i=0;i<N;i++) {
			if(s.charAt(i)!=now) {
				list.add(i-begin);
				now=s.charAt(i);
				begin=i;
			}
		}
		list.add(N-begin);
		if(now == '0')	list.add(0);

		/*
		for(int i:list) System.out.print(i+" ");
		System.out.println();
		 */

		int[] sum=new int[list.size()+1];
		for(int i=0;i<list.size();i++) {
			sum[i+1]=sum[i]+list.get(i);
		}

		int result = 0;


		for(int i = 0; i < list.size(); i+=2) {

			int Left = i;
			int Right = Math.min(i+K*2+1, list.size());

			int tmp=sum[Right]-sum[Left];

			//System.out.println(String.format("left:%d right:%d tmp:%d",Left,Right,tmp));

			result = Math.max(tmp, result);
		}

		System.out.println(result);


		sc.close();
	}

}
