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 P608E {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] line = br.readLine().split(" ");
        long n = Long.parseLong(line[0]);
        long k = Long.parseLong(line[1]);


        long min = 1;
        long max = n/2 + 1;

        while(min != max) {
            long mid = min + (max - min + 1)/2;

            long paths = calcPaths(2*mid, n);

            if(paths >= k) {
                min = mid;
            } else {
                max = mid-1;
            }
        }

        long even = calcPaths(2*min, n);
        long oddTwo = calcPaths(2*min+1, n);

        if(oddTwo >= k)
            System.out.println(2*min+1);
        else if(even >= k)
            System.out.println(2*min);
        else
            System.out.println(2*min-1);


    }

    public static long calcPaths(long val, long max) {
        if(val == 1 || val == 2)
            return max + 1 - val;
        else {
            long count = 0;
            long oneWidth = (val & 1) == 1 ? 1 : 2;
            long zeroWidth = val - oneWidth;
            for(long cur = val; cur<=max;) {
                if(cur - 1 + oneWidth > max) {
                    count += max - cur + 1;
                    cur = max + 1;
                } else {
                    count += oneWidth;
                    cur += oneWidth + zeroWidth;
                    oneWidth *= 2;
                    zeroWidth *= 2;
                }
            }
            return count;
        }
    }
}
