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 {
	ArrayList<Integer> t = new ArrayList<Integer>();
	int count;
	String num = "1";
	int sum;
	int max =1;

	Main(int n,int k,String[] S){
		for(int p=0;p<n;p++){
			if(!(S[p].equals(num))){
				if(num.equals("0")){
					num="1";
				}else{
					num="0";
				}
				t.add(count);
				count=0;
			}
			count++;
		}

		t.add(count);			//ラス1回追加
		if(num.equals("0"))t.add(0);	//10並びにしたいので、終端数が1の場合は末尾0が0個扱い
      
		Iterator it = t.iterator();
		count=0;
		while(it.hasNext()&&count<k*2+1){
			sum += (int)(it.next());
			count++;
		}

		max=sum;

		while(count<t.size()){
			sum += t.get(count);
			sum += t.get(count+1);
			sum -= t.get(count-k*2-1);
			sum -= t.get(count-k*2);
			if(sum>max)max=sum;
			count+=2;
		}

		// 出力
		System.out.println(max);
	}

	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();

		String s = sc.next();
		String[] S = s.split("");

		new Main(n,k,S);
	}
}