E. Common Numbertime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputAt 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} 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 \le x \le n, y \in path(x) \} \right| \ge k.InputThe first line contains two integers n and k (1 \le k \le n \le 10^{18}).OutputPrint the only integer — the maximum value that is contained in at least k paths.ExamplesInput
11 3
Output
5
Input
11 6
Output
4
Input
20 20
Output
1
Input
14 5
Output
6
Input
1000000 100
Output
31248
NoteIn 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.
11 3

5 
['binary search', 'combinatorics', 'dp', 'math', '*2100']