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 {
	
	static final int INF = Integer.MAX_VALUE/2;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), k = sc.nextInt();
		String s = sc.next();
		char c[] = s.toCharArray();
		int a[] = new int[n];
		for(int i=0;i<n;i++) a[i] = c[i] - '0';
		
		int now = 1;
		int cnt = 0;
		List<Integer> ls = new ArrayList<>();
		for(int i=0;i<n;i++) {
			if(now == a[i]) {
				cnt++;
			}
			else {
				ls.add(cnt);
				cnt = 1;
				now = 1 - now;
			}
		}
		ls.add(cnt);
		
		if(a[n-1] == 0) ls.add(0);
				
		int len = ls.size();
		
		if(len/2<=k) {
			System.out.println(n);
			return;
		}
		
		int sum[] = new int[len+1];
		for(int i=0;i<len;i++) sum[i+1] = sum[i] + ls.get(i);
		
		int ans = 0;
		for(int i=0;i<len;i+=2) {
			if(i + k*2 + 1 >= len + 1)break;
			ans = Math.max(ans,sum[i + k*2 + 1] - sum[i]);			
		}
		System.out.println(ans);
	}
}

