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[] args) {
		Main main = new Main();
		main.run();
	}

	public void run() {
		Scanner sc = new Scanner(System.in);
		int N=sc.nextInt();
		int K=sc.nextInt();
		char S[] = sc.next().toCharArray();

		ArrayList<Integer>Nums = new ArrayList<>();
		int now = 1;
		int cnt = 0;
		for(int i=0; i<N; i++) {
			if(S[i] == (char)('0' + now)) {
				cnt++;
			}else {
				Nums.add(cnt);
				now = 1 - now;
				cnt = 1;
			}
		}
		if(cnt>0) Nums.add(cnt);

		if(Nums.size()%2==0) Nums.add(0);

		ArrayList<Integer>sums = new ArrayList<Integer>();
		sums.add(0);
		for(int i=0; i<Nums.size(); i++) {
			sums.add(i+1, sums.get(i) + Nums.get(i));
		}

		int add = 2 * K + 1;
		int ans = 0;
		for(int i=0; i<Nums.size(); i+=2) {
			int left = i;
			int right = Math.min(i + add, Nums.size());
			ans = Math.max(ans, sums.get(right) - sums.get(left));
		}
		System.out.println(ans);
		sc.close();
	}

}
