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) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		char[] S = sc.next().toCharArray();
		int ans1 = 0, ans = 0;
		List<Integer> a = new ArrayList<>();
		int con = 1;
		if(N == 1)
			a.add(1);
		for(int i = 1; i < N; i++) {
			if(S[i] == S[i-1]) {
				con += 1;
			}
			else {
				a.add(con);
				con = 1;
			}
			if(i == N-1)
				a.add(con);
		}
		if(S[0] == '1') {
			LOOP:for(int i = 0; i < a.size(); i += 2) {
				ans1 = 0;
				for(int j = i; j <= i + K*2 && j < a.size(); j++) {
					ans1 += a.get(j);
				}
				ans = Math.max(ans, ans1);
				if(i + K*2 >= a.size())
					break LOOP;
			}
		}
		else {
			for(int i = 0; i < K*2 && i < a.size(); i++)
				ans += a.get(i);
			LOOP:for(int i = 1; i < a.size(); i += 2) {
				ans1 = 0;
				for(int j = i; j <= i + K*2 && j < a.size(); j++) {
					ans1 += a.get(j);
				}
				ans = Math.max(ans, ans1);
				if(i + K*2 >= a.size())
					break LOOP;
			}
		}
		System.out.println(ans);
	}

}