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.Arrays;
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();
		sc.close();

		int[] start = new int[N+1];
		int[] end = new int[N+1];

		Arrays.fill(start, 999999);
		Arrays.fill(end, 999999);

		int scount = 0;
		int ecount = 0;

		if(S.substring(0, 1).equals("0")){
			start[0] = 0;
			scount++;
		}
		for(int i = 0 ; i < S.length()-1; i++){
			if(S.substring(i, i+1).equals("1") && S.substring(i+1,i+2).equals("0")){
				start[scount] = i+1;
				scount++;
			}
			if(S.substring(i, i+1).equals("0") && S.substring(i+1,i+2).equals("1")){
				end[ecount] = i;
				ecount++;
			}
		}

		int ans = 0;
		int count = 0;
		for(int i = 0 ; i < S.length(); i++){
			int p = start[K+count];
			if(start[K+count] == 999999){
				p = S.length();
			}
			//System.out.println(p-i);
			if(ans < p-i){
				ans = p-i;
			}

			if(end[count] == 999999){
				break;
			}

			i = end[count];
			count++;

		}

		System.out.println(ans);

	}
}
