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> group = new ArrayList<Integer>();
		int len = 0; char c = S.charAt(0); int sum = 0; int max = 0; int ML = 2 * k + 1;

		for(int i = 0; i < n; i++) {
			if(i == 0 && c != '1') group.add(0);
			if(c == S.charAt(i)) {
				len++;
			}else {
				group.add(len);
				len = 1;
				c = (c == '0') ? '1' : '0';
			}
			if(i == n - 1) {
				if(c == '1') {
					group.add(len);
				}else {
					group.add(len);
					group.add(0);
				}
			}
		}

		if(group.size() < ML) {
			for(int i : group) max += i;
		}else {
			for(int i = 0; i < ML;  i++) sum += group.get(i);
			max = sum;
			for(int i = 2; i + ML - 1 < group.size(); i += 2) {
				sum = sum + group.get(i + ML -1) + group.get(i + ML -2) - group.get(i-1) - group.get(i-2);
				max = Math.max(sum, max);
			}
		}
		System.out.println(max);

		sc.close();
	}

}