At first, let's define function f(x) as follows: $$$ \begin{matrix} f(x) & = & \left\{ \begin{matrix} x/2 & \mbox{if } x  is even \\\ x - 1 & \mbox{otherwise } \end{matrix} \right. \end{matrix} $$$

We can see that if we choose some value v and will apply function f to it, then apply f to f(v), and so on, we'll eventually get 1. Let's write down all values we get in this process in a list and denote this list as path(v). For example, path(1) = [1], path(15) = [15, 14, 7, 6, 3, 2, 1], path(32) = [32, 16, 8, 4, 2, 1].

Let's write all lists path(x) for every x from 1 to n. The question is next: what is the maximum value y such that y is contained in at least k different lists path(x)?

Formally speaking, you need to find maximum y such that \left| \{ x ~|~ 1 ≤ x ≤ n, y ∈ path(x) \} \right| ≥ k.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 10^{18}).

Output

Print the only integer — the maximum value that is contained in at least k paths.

Examples

Input


11 3


Output


5


Input


11 6


Output


4


Input


20 20


Output


1


Input


14 5


Output


6


Input


1000000 100


Output


31248

Note

In the first example, the answer is 5, since 5 occurs in path(5), path(10) and path(11).

In the second example, the answer is 4, since 4 occurs in path(4), path(5), path(8), path(9), path(10) and path(11).

In the third example n = k, so the answer is 1, since 1 is the only number occuring in all paths for integers from 1 to 20.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author shangu
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        ECommonNumber solver = new ECommonNumber();
        solver.solve(1, in, out);
        out.close();
    }

    static class ECommonNumber {
        private long n;

        public void solve(int testNumber, FastReader in, PrintWriter out) {
            n = in.nextLong();
            long k = in.nextLong();
            long L = 0;
            long R = n / 2 + 1;
            while (R - L > 1) {
                long M = (L + R) / 2;
                if (count(M * 2) >= k) L = M;
                else R = M;
            }
            long ans = L * 2;
            L = 0;
            R = (n - 1) / 2 + 1;
            while (R - L > 1) {
                long M = (L + R) / 2;
                if (count(4 * M + 2) + 1 >= k) L = M;
                else R = M;
            }
            ans = Math.max(ans, L * 2 + 1);
            out.println(ans);
        }

        private long count(long x) {
            long cnt = 0, curcnt = 2;
            while (x <= n) {
                cnt += Math.min(curcnt, n - x + 1);
                x <<= 1;
                curcnt <<= 1;
            }
            return cnt;
        }

    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream), 32768);
            st = null;
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

    }
}

