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.util.*;
import java.io.*;
public class EdE {

	public static void main(String[] args) throws Exception{
		long num = 998244353;

		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		PrintWriter out = new PrintWriter(System.out);
 		StringTokenizer st = new StringTokenizer(bf.readLine());
 		long n = Long.parseLong(st.nextToken());
 		long k = Long.parseLong(st.nextToken());
 		long l = 1;
 		long r = (n+1L)/2L;
 		while(l < r){
 			long mid = (l+r+1L)/2L;
 			if (appearFirstN(n, 2L*mid) >=k)
 				l = mid;
 			else
 				r = mid-1;
 		}
 		
 		if (n == k)
 			out.println(1);
 		else
 			out.println(appearFirstN(n, 2L*l+1L) >= k ? 2L*l+1L : 2L*l);
	 		
 		out.close();
 		
 		
 		
 	}
	public static long appearFirstN(long n, long k){
		if (k > n)
			return 0;
		long ans =0;
		if (k%2 == 0){
			k/=2;
			ans-=1;
		}
		
		String a = Long.toString(n, 2);
		long prod = 1;
		String b = Long.toString(k, 2);
		if (a.substring(0, b.length()).equals(b)){
			ans+= (Long.parseLong("0" + a.substring(b.length()), 2)+1);
		}
		else if (a.substring(0, b.length()).compareTo(b) > 0){
			prod = 2;
		}
		
		for(int j =0;j<a.length()-b.length();j++){
			prod*=2L;
		}
		ans+= (prod-1);
		return ans;
		
	}
}
 	
 
//StringJoiner sj = new StringJoiner(" "); 
//sj.add(strings)
//sj.toString() gives string of those stuff w spaces or whatever that sequence is

 		
 		
 		
 		
	


