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 Quarantine;

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

public class CommonNumber {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        long n=Long.parseLong(st.nextToken());
        long k=Long.parseLong(st.nextToken());
        long anso=1,anse=0;
        long low=1,high=(n+1)/2;
        while(low<=high){
            long mid=(low+high)/2;
            long m=2*mid-1;
            if(getAns(n,m)>=k){
                anso=m;
                low=mid+1;
            }
            else{
                high=mid-1;
            }
        }
        low=1;
        high=n/2;
        while(low<=high){
            long mid=(low+high)/2;
            long m=2*mid;
            if(getAns(n,m)>=k){
                anse=m;
                low=mid+1;
            }
            else{
                high=mid-1;
            }
        }
        System.out.println(Math.max(anse,anso));
    }
    public static long getAns(long n,long m){
        long l=m,r=m;
        if(m%2==0){
            r++;
        }
        long ans=0;
        while(r<=n){
            ans+=r-l+1;
            l=2*l;
            r=2*r+1;
        }
        if(l<=n)
        ans+=n-l+1;
//        System.out.println(n+" "+m+" "+ans);
        return ans;
    }
}
