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.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;

public class Main {
    
    public static void main(String[] args) throws IOException {
        
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        String[] str = bf.readLine().split(" ");
        int N = Integer.parseInt(str[0]), K = Integer.parseInt(str[1]), max = 0;
        String[] S = bf.readLine().split("");
        ArrayList<Integer> data = new ArrayList<>();
        int value = Integer.parseInt(S[0]);
        if (value == 0) {
            data.add(0);
        }
        for (int i = 0; i < N; value = 1 - value) {
            int count = 0;
            while (i < N && Integer.parseInt(S[i]) == value) {
                count++;
                i++;
            }
            data.add(count);
        }
        if (value == 1) {
            data.add(0);
        }
        int size = data.size(), sum = 0;
        for (int i = 0; i < 2 * K + 1 && i < size; i++) {
            sum += data.get(i);
        }
        max = sum;
        for (int i = 3; i + 2 * K - 1 < size; i += 2) {
            sum += -data.get(i - 3) - data.get(i - 2) + data.get(i + 2 * (K - 1)) + data.get(i + 2 * K - 1);
            if (max < sum) {
                max = sum;
            }
        }
        System.out.println(max);
    }
}