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.Vector;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		char s[] = sc.next().toCharArray();

		char target[] = {'1', '0'};
		int targetIndex = 0;
		Vector<Integer> vec = new Vector<Integer>();
		int count = 0;
		for(int i = 0; i < n; i++){

			if(s[i] != target[targetIndex]){
				targetIndex++;
				targetIndex %= 2;
				vec.add(count);
				count = 0;
			}

			count++;

			if(i == n - 1){
				vec.add(count);
				break;
			}
		}

		for(int i = 0; i < vec.size(); i++){
//			System.out.println(vec.get(i));
		}

		int sum = 0;
		for(int i = 0; i < Math.min(vec.size(), 2*k + 1); i++){
			sum += vec.get(i);
		}
//		System.out.println("sum = "+sum);

		int tmpSum = sum;
		for(int i = 2; i < vec.size(); i += 2){
			tmpSum -= (vec.get(i - 1) + vec.get(i - 2));
//			System.out.println("remove "+(i - 2)+ " and " + (i - 1));
			if(i + 2*k - 1 < vec.size()){
				tmpSum += vec.get(i + 2*k - 1 );
//				System.out.println("add "+(i + 2*k - 1));
			}
			if(i + 2*k  < vec.size()){
				tmpSum += vec.get(i + 2*k  );
//				System.out.println("add "+(i + 2*k));
			}

//			System.out.println("tmpSum "+tmpSum);
			sum = Math.max(sum, tmpSum);
		}

		System.out.println(sum);
	}

}