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.*;

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();

		//zyunbi
		int temp=0;
		int now=1;
		List<Integer> list = new ArrayList<Integer>();
		for (int i=0;i<N;i++) {
			if (S.charAt(i)-48==0+now) {
				temp++;
			} else {
				list.add(temp);
				now=1-now;
				temp=1;
			}
		}
		if (temp!=0) {
			list.add(temp);
		}
		if (list.size()%2==0) {
			list.add(0);
		}

		//ruiseki
		/*
		List<Integer> R = new ArrayList<Integer>();
		int X=0;
		R.add(0);
		for (int i=0;i<list.size();i++) {
			X=list.get(i)+R.get(i);
			R.add(X);
		}

		int add=2*K+1;
		int left;
		int right;
		int temp_A;
		int max=0;
		for (int i=0;i<R.size();i+=2) {
			left=i;
			right=Math.min(left+add,R.size()-1);
			temp_A=R.get(right)-R.get(left);
			max=Math.max(max,temp_A);
		}
		*/

		//syakutori
		int add=2*K+1;
		int left=0;
		int right=0;
		int tmp=0;
		int max=0;
		for (int i=0;i<list.size();i+=2) {
			int Nextleft=i;
			int Nextright=Math.min(i+add,list.size());
			while (Nextleft>left) {
				tmp-=list.get(left);
				left++;
			}

			while (Nextright>right) {
				tmp+=list.get(right);
				right++;
			}
			max=Math.max(max,tmp);
		}
		System.out.println(max);
	}
}