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);

		int result = 0;
		int sum = 0;
		int left = 0;
		int right = 0;
		for(int i = 0; i < list.size(); i+=2) {
			int nextLeft = i;
			int nextRight = Math.min(i+k*2+1, list.size());
			
			while(nextLeft > left) {
				sum -= list.get(left);
				left++;
			}
			
			while(nextRight > right) {
				sum += list.get(right);
				right++;
			}
			result = Math.max(sum, result);
		}

		System.out.println(result);
	}
}