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.
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner s = new Scanner(System.in);
		long n = s.nextLong();
		long k = s.nextLong();
		if(n==k){
		    System.out.println(1);
		    return;
		}
        long i=1;
        long t1=1;
        long t2=2;
        if(n%2==0) t2=1;
        while(2*i<k){
            if(n%2==0){
                if(n%4==0){
                    t1=i+1+t1;
                }
                else{
                    t1=t1+1;
                }
            }
            else{
                if((n-1)%4==0){
                    t1=i+1+t2;
                }
                else{
                    t1=t2+1;
                }
            }
            n/=2;
            if(n%2!=0) t2 = t1 + 2*i + 1;
            else t2 = 2*i+1;
            i=2*i+1;
        }
        if(t1>=k){
            System.out.println(n);
            return;
        }
        if(n%2==0) {
            if(i>=k) System.out.println(n-1);
            else
            System.out.println(n-2);
        }
        else{
            if(t2>=k)
            System.out.println(n-1);
            else{
                if(i>=k) System.out.println(n-2);
                else
                System.out.println(n-3);
            }
        }
	}
}
