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

        int N = sc.nextInt();
        int K = sc.nextInt();
        sc.nextLine();
        String S = sc.nextLine();

        List<Integer> switches = new ArrayList<>();
        switches.add(0);
        boolean currentIsReverse = (S.charAt(0) == '1');

        for (int i = 1; i < N; i++) {
            boolean isReverse = (S.charAt(i) == '1');
            if (isReverse != currentIsReverse) {
                switches.add(i);
            }
            currentIsReverse = isReverse;
        }

        int numOfSwitch = switches.size();
        int maxSum = 0;


        for (int i = 0; i < numOfSwitch; i++) {
            boolean isFirstReverse = (S.charAt(switches.get(i)) == '1');

            int endIndex = i + 2 * K + (isFirstReverse ? 1 : 0);
            int sum = ((endIndex < numOfSwitch) ? switches.get(endIndex) : N) - switches.get(i);

            if (maxSum < sum) maxSum = sum;
        }

        System.out.println(maxSum);

    }
}
