A13129. Common Number

At first, let's define function $f(x)$ as follows: $$$$ \begin{matrix} f(x) & = & \left\{ \begin{matrix} \frac{x}{2} & \mbox{if } x \text{ is even} \\ x - 1 & \mbox{otherwise } \end{matrix} \right. \end{matrix} $$ </p><p>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) = \[32, 16, 8, 4, 2, 1\]$ .</p><p>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 $path(x)$ ?</p><p>Formally speaking, you need to find maximum $y$ such that $\\left| \\{ x ~|~ 1 \\le x \\le n, y \\in path(x) \\} \\right| \\ge k$$$.

The first line contains two integers $n$ and $k$ ( $1 \le k \le n \le 10^{18}$ ).

Print the only integer — the maximum value that is contained in at least $k$ paths.

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 $20$ .