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.math.BigDecimal;
import java.util.Arrays;
import java.util.Scanner;
import java.util.Vector;

public class Main {


    static class MyPair{
        int k,v;
        MyPair(int first,int second) {
            this.k = first;
            this.v = second;
        }
    }
    public static void main(String[] args) {
        java.util.Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int k = input.nextInt();
        String s = input.next();
        int[] ss = new int[n];
        Vector<MyPair> d = new Vector<>();
        int prev = -1, cnt = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '0') ss[i] = 0;
            else ss[i] = 1;
        }
        for (int i = 0; i < n; i++) {
            if (prev == ss[i]) cnt++;
            else {
                if (cnt > 0) d.add(new MyPair(prev, cnt));
                cnt = 1;
                prev = ss[i];
            }
        }
        if (cnt > 0) d.add(new MyPair(prev, cnt));
        int l = 0, r = 0, sum = 0, v = 0;
        for (int i = 0; i < d.size(); i++) {
            sum += d.get(i).v;
            if (d.get(i).k == 0) {
                v++;
                while (v > k) {
                    sum -= d.get(l).v;
                    if (d.get(l).k == 0) v--;
                    l++;
                }
            }
            if (r < sum) r = sum;
        }
        System.out.println(r);
    }
}
