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.util.Scanner;

public class e {
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		//		int n=11, t=6;
		//		int ct=0;
		//		for(int i=1;i<=n;i++) {
		//			int x=i;
		//			while(x>1) {
		//				if(x%2==0) x/=2;
		//				else x--;
		//				if(x==t||i==t) {
		//					System.out.println(i);
		//					ct++;
		//					break;
		//				}
		//			}
		//		}
		//		System.out.println(ct);

		long n=scan.nextLong(), k=scan.nextLong();
		if(n==k) {
			System.out.println(1);
			return;
		}

		long res=-1;
		long lo=1, hi=n/2;

//		System.out.println(good(40,n,k));
		while(hi>=lo) {
			long mid=(hi+lo)/2;
			if(good(mid*2,n,k)) {
				res=Math.max(res,mid*2);
				lo=mid+1;
			}
			else {
				hi=mid-1;
			}
		}

		lo=0;
		hi=n/2;

		while(hi>=lo) {
			long mid=(hi+lo)/2;
			if(good(mid*2+1,n,k)) {
				res=Math.max(res,mid*2+1);
				lo=mid+1;
			}
			else {
				hi=mid-1;
			}
		}

		System.out.println(res);
	}

	public static boolean good(long x, long n, long k) {
		long c=0L;
		long cu=x;
		long p=2-x%2;

		while(true) {
			if(cu+p-1>n) {
				c+=Math.max((p+n-(cu+p-1)),0);
				break;
			}
			cu*=2;
			c+=p;
			p*=2;
		}
		return c>=k;
	}
}