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 Div2608;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class ProblemE {
    public static long getCount(long m,long n){
        long left,right;
        long count=0;
        if(m%2==0){
            left=m;
            right=m+1;
        }
        else{
            left=right=m;
        }
        while(true){
            if(n>=left&&n>=right){
                count+=right-left+1;
                left=2*left;
                right=2*right+1;
            }
            else if(n>=left&&n<right){
                count+=n-left+1;
                break;
            }
            else{
                break;
            }
        }
        return count;
    }
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        String line[]=br.readLine().split(" ");
        long n=Long.parseLong(line[0]);
        long k=Long.parseLong(line[1]);
        long left=1;
        long right=n;
        long ans=1;
//        long ansevven=1;
        while(left<=right){
            long mid=(left+right)/2;
            long m=mid*2;
            long temp=getCount(m,n);
//            System.out.println(mid+" "+temp);
            if(temp>=k){
                left=mid+1;
                ans=Math.max(ans,m);
            }
            else{
                right=mid-1;
            }
        }
        left=1;right=n;
        while (left<=right){
            long mid=(left+right)/2;
            long m=mid*2-1;
            long temp=getCount(m,n);
//            System.out.println(mid+" "+temp);
            if(temp>=k){
                left=mid+1;
                ans=Math.max(ans,m);
            }
            else{
                right=mid-1;
            }
        }
        System.out.println(ans);
    }
}
