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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        char[] s = sc.next().toCharArray();
        List<Integer> one = new ArrayList<>();
        List<Integer> zero = new ArrayList<>();
        int oneCnt = 0;
        int zeroCnt = 0;
        if (s[0] == '1') {
            oneCnt++;
        } else {
            zeroCnt++;
            one.add(0);
        }
        for (int i = 1; i < n; i++) {
            if (s[i] == '1') {
                oneCnt++;
                if (zeroCnt != 0) {
                    zero.add(zeroCnt);
                    zeroCnt = 0;
                }
            } else {
                zeroCnt++;
                if (oneCnt != 0) {
                    one.add(oneCnt);
                    oneCnt = 0;
                }
            }
        }
        if (oneCnt != 0) {
            one.add(oneCnt);
        } else if (zeroCnt != 0) {
            zero.add(zeroCnt);
            one.add(0);
        }

        if (zero.isEmpty() || zero.size() == 1 && one.get(0) == 0 && one.get(1) == 0) {
            System.out.println(s.length);
            return;
        }
        long oneSum = 0;
        long zeroSum = 0;
        for (int i = 0; i < Math.min(k + 1, one.size()); i++) {
            oneSum += one.get(i);
        }
        for (int i = 0; i < Math.min(k, zero.size()); i++) {
            zeroSum += zero.get(i);
        }
        long ans = zeroSum + oneSum;
        for (int i = 1; i < zero.size() - k + 1; i++) {
            zeroSum += zero.get(i + k - 1) - zero.get(i - 1);
            oneSum += one.get(i + k) - one.get(i - 1);
            ans = Math.max(zeroSum + oneSum, ans);
        }
        System.out.println(ans);
    }

}
