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.io.*;
import java.util.*;
 
public class Main {
	static class FastReader{ 
        BufferedReader br; 
        StringTokenizer st; 
        public FastReader(){br = new BufferedReader(new InputStreamReader(System.in));}
        String next(){while (st == null || !st.hasMoreElements()){try{st = new StringTokenizer(br.readLine());} 
        catch (IOException  e){e.printStackTrace();}}return st.nextToken();}
        int nextInt(){ return Integer.parseInt(next());}long nextLong(){return Long.parseLong(next());}double nextDouble(){return Double.parseDouble(next());}
        String nextLine(){String str = ""; try{str = br.readLine(); } catch (IOException e) {e.printStackTrace();} return str; } 
    }
	static FastReader sc = new FastReader();
	static long mod = (long)(1e9+7);
	public static void main (String[] args){
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;
//        t = sc.nextInt();
        z :while(t-->0){
        	long n = sc.nextLong();
           	long k = sc.nextLong();
           	
           	if(n == k) {
           		out.write(1+"\n");
           		continue;
           	}
           	long res = -1;
           	long l = 0, h = n/2;
           	while(l<=h) {
           		long mid = (l+h)/2;
           		if(cal(mid*2,n,k)) {
           			res = Math.max(mid*2, res);
           			l = mid+1;
           		}
           		else h = mid-1;
           	}
           	l = 1; h = n/2;
           	while(l<=h) {
           		long mid = (l+h)/2;
           		if(cal(mid*2+1,n,k)) {
           			res = Math.max(mid*2+1, res);
           			l = mid+1;
           		}
           		else h = mid-1;
           	}
           	out.write(res+"\n");
       	}
		out.close();
	}
	private static boolean cal(long l, long n, long k) {
		long cnt = 0L;
		long cur = l;
		long add = 2 - l%2L;
		while(true) {
			if(l+add-1>n) {
				cnt += Math.max(n-l+1, 0);
				break;
			}
			cnt += add;
			add*=2;
			l*=2;
		}
		return cnt>=k;
	}
}
