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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        int K = scan.nextInt();
        String S = scan.next();
        int answer = solve(N, K, S);
        System.out.println(answer);
    }
    public static int solve(int N, int K, String S) {
        char[] data = S.toCharArray();
        List<Integer> group = new ArrayList<>();
        List<Integer> kind = new ArrayList<>();
        int length = data.length;
        char prev = ' ';
        int count = 0;
        for (int i = 0; i < length; i++) {
            if (prev != data[i]) {
                if (count != 0) {
                    group.add(count);
                    kind.add(prev == '0' ? 1 : 0);
                }
                prev = data[i];
                count = 1;
                continue;
            }
            count += 1;
        }
        if (0 < count) {
            group.add(count);
            kind.add(prev == '0' ? 1 : 0);
        }
        Integer[] group_array = group.toArray(new Integer[0]);
        Integer[] kind_array = kind.toArray(new Integer[0]);
        int len = group_array.length;
        int[] g = new int[len + 1];
        int[] k = new int[len + 1];
        Arrays.fill(g, 0);
        Arrays.fill(k, 0);
        for(int i = 0; i < len; i++) {
            g[i+1] += g[i] + group_array[i];
            k[i+1] += k[i] + kind_array[i];
        }
        int answer = 0;
        int j = 1;
        for(int i = 1; i <= len; i++) {
            if (j < i) {
                j = i;
            }
            for(; j <= len; j++) {
                if (k[j] - k[i-1] <= K) {
                    answer = Math.max(answer, g[j] - g[i - 1]);
                    continue;
                }
                break;

            }
        }
        return answer;
    }
}
