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.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
import java.util.StringTokenizer;

public class E {

	public static void main(String[] args) throws Exception {
//		StringTokenizer stok = new StringTokenizer(new Scanner(new File("F:/books/input.txt")).useDelimiter("\\A").next());
		StringTokenizer stok = new StringTokenizer(new Scanner(System.in).useDelimiter("\\A").next());
		StringBuilder sb = new StringBuilder();
		long n = Long.parseLong(stok.nextToken());
		long k = Long.parseLong(stok.nextToken());
		if(n==k) System.out.println("1");
		else if(k==n-1) System.out.println("2");
		else if(k==1) System.out.println(n);
		else {
			long v1 = solveEven(n,k);
			long v2 = solveEven(n,k-1)/2;
			System.out.println(Math.max(v1, v2));
		}
//		System.out.println("******");
//		System.out.println(getEven(232830644l,1000000000000000000l));
//		System.out.println(getEven(232830642l,1000000000000000000l));
//		System.out.println(getEven(232830640l,1000000000000000000l));
	}

	private static long solveEven(long n, long k) {
		long lo = 2;
		long hi = n-n%2;
		while(lo!=hi) {
			long m = (lo+hi)/2;
			m = m+m%2;
			long v = getEven(m,n);
//			System.out.println(m+" "+v +" "+lo+" "+hi);
			if(v<k) hi = m-2; else lo=m;
		}
		return lo;
	}

	private static long getEven(long m, long n) {
		long cur=2;
		long tot = 0;
		while(m+cur-1<=n) {
			tot+=cur;
//			System.out.println(m+" "+tot+" "+cur);
			m*=2;
			cur*=2;
		}
		if(m<=n) {
			tot += n-m+1;
//			System.out.println(m+" "+tot+" "+cur);
		}
		
		return tot;
	}

}
