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 long N;
	static long countprefix(long num) {
	    int zero_bits = 0;
	    long count=0;
	    while (num<= N) {
	        count += Math.min(N+1,num+(1L<<zero_bits))-num;
	        num *= 2;
	        zero_bits++;
	    }
	    return count;
	}
	static long countpaths(long n) {
	    long count = countprefix(n);
	    if (n%2==0)
	        count+=countprefix(n+1);
	    return count;
	}
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int ttt=1;
//		ttt=f.nextInt();
		PrintWriter out=new PrintWriter(System.out);
		for(int tt=0;tt<ttt;tt++) {
			N=f.nextLong();
			long k=f.nextLong();
			long ans=0;
			for (int bit = 59; bit >= 0; bit--)
		        if (countpaths(ans+(1L<< bit))>=k)
		        	ans += 1L<< bit;
			System.out.println(ans);
		}
		out.close();
	} 
	static void sort(int[] p) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: p) q.add( i);
        Collections.sort(q);
        for (int i = 0; i < p.length; i++) p[i] = q.get(i);
    }
    
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 	