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.Scanner;
import java.util.ArrayList;
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();
		ArrayList<Integer> zero=new ArrayList<>();
		ArrayList<Integer> one=new ArrayList<>();
		boolean ch=false;
		boolean ii=false;
		if(s.charAt(0)=='1'){
			ch=false;
			ii=false;
		}else{
			ch=true;
			ii=true;
		}
		for(int i = 0; i<n; i++){
			if(s.charAt(i)=='1' && !ch){
				one.add(i);
				ch=true;
			}else if(s.charAt(i)=='0' && ch){
				zero.add(i);
				ch=false;
			}
		}
		if(zero.size()==0){
			System.out.println(s.length());
			return;
		}
		int ans=0;
		int sub=0;
		for(int i=0; i<=zero.size()-k; i++){
			if(k+i<zero.size()){
				ans=Math.max(ans,zero.get(k+i)-sub);
				if(ii){
					sub=one.get(i);
				}else{
					sub=one.get(i+1);
				}
			}else{
				ans=Math.max(ans,s.length()-sub);
			}
		}
		if(k>zero.size()){
			ans=s.length();
		}
		System.out.println(ans);
	}
}
