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 static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.System.exit;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {
	static long n, m;
	static void solve() throws Exception {
		n = nextLong();
		m = nextLong();
		long res = 1;
		long cn = n;
		while (cn!=0) {
			for (long i = 0; i <= 10; i++) {
				if (i >= cn) {
					continue;
				}
				if (vtext(cn - i) >= m) {
					res = max(res, cn - i);
					break;
				}
			}
			cn /= 2;
		}
		System.out.println(res);

	}

	private static long vtext(long l) {
		if (l % 2 == 0) 
			return test(l) + test(l + 1);
		
		return test(l);
	
	}
	private static long test(long l) {
		long p2 = 1;
		long ans = 0;
		while (p2 * l <= n) {
			long h = min(n, p2 * (l + 1) - 1);
			ans += h - p2 * l + 1;
			p2 *= 2;
		}
		return ans;
	}

	static int nextInt() throws IOException {
		return parseInt(nextString());
	}

	static long nextLong() throws IOException {
		return parseLong(nextString());
	}

	static String nextString() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;

	public static void main(String[] args) {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve();
			in.close();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			exit(1);
		}
	}
}