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;

public class Main {

    public static void main(final String[] args) {
        try (final Scanner sc = new Scanner(System.in)) {
            final int n = sc.nextInt();
            final int k = sc.nextInt();
            final String s = sc.next();

            final char first = s.charAt(0);
            final int[] cumulatives = new int[n + 1];
            int cSize = 0;
            cumulatives[cSize++] = 0;
            if (first == '0') {
                cumulatives[cSize++] = 0;
            }
            char prev = first;
            int len = 0;
            for (int i = 0; i < n; i++) {
                final char current = s.charAt(i);
                if (current != prev) {
                    cumulatives[cSize] = cumulatives[cSize - 1] + len;
                    cSize++;
                    len = 1;
                } else {
                    len += 1;
                }
                prev = current;
            }
            if (len > 0) {
                cumulatives[cSize] = cumulatives[cSize - 1] + len;
                cSize++;
            }
            int answer = 0;
            for (int left = 0, right = Math.min(left + 2 * k + 1, cSize - 1); left < cSize; left += 2, right = Math
                    .min(left + 2 * k + 1, cSize - 1)) {
                final int value = cumulatives[right] - cumulatives[left];
                answer = Math.max(answer, value);
            }
            System.out.println(answer);
        }
    }

}
