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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        char[] c = sc.next().toCharArray();
        List<Integer> nums = new ArrayList<>();
        char now = '1';
        int count = 0;
        for (int i = 0 ; i < n ; i++) {
            if (c[i] == now) {
                count++;
            } else {
                nums.add(count);
                if (now == '0') {
                    now = '1';
                } else {
                    now = '0';
                }
                count = 1;
            }
        }
        nums.add(count);
        if (nums.size() % 2 == 0) {
            nums.add(0);
        }

        int[] a = new int[nums.size() + 1];
        int sum = 0;
        for (int i = 0 ; i < nums.size() ; i++) {
            sum += nums.get(i);
            a[i + 1] = sum;
        }

        int add = Math.min(nums.size(), 2 * k + 1);
        int max = 0;
        for (int i = 0 ; i < nums.size() + 1 ; i += 2) {
            if (i + add < nums.size() + 1) {
                max = Math.max(max, a[i + add] - a[i]);
            }
        }
        if (nums.size() == 1) {
            System.out.println(n);
        } else {
            System.out.println(max);
        }

    }

}
