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.ArrayDeque;
import java.util.Deque;
import java.util.Scanner;

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

        sc.nextInt();
        int K = sc.nextInt();
        String S = sc.next();

        System.out.println(solve(S, K));

        sc.close();
    }

    static int solve(String S, int K) {
        int result = -1;
        Deque<Element> deque = new ArrayDeque<>();
        int totalLength = 0;
        int directionNum = 0;
        for (char value : S.toCharArray()) {
            if (!deque.isEmpty() && deque.peekLast().value == value) {
                ++deque.peekLast().length;
                ++totalLength;
            } else {
                if (value == '0') {
                    ++directionNum;
                }
                deque.offerLast(new Element(value, 1));
                ++totalLength;

                while (directionNum == K + 1) {
                    Element head = deque.pollFirst();
                    totalLength -= head.length;
                    if (head.value == '0') {
                        --directionNum;
                    }
                }
            }

            result = Math.max(result, totalLength);
        }

        return result;
    }
}

class Element {
    char value;
    int length;

    Element(char value, int length) {
        this.value = value;
        this.length = length;
    }
}