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.
#include <bits/stdc++.h>
using namespace std;
long long po[61];
int main() {
  po[0] = 1;
  for (int i = 1; i < 61; i++) po[i] = po[i - 1] * 2;
  long long n, a;
  cin >> n >> a;
  if (n == a) {
    cout << 1;
    return 0;
  }
  if (n == a + 1) {
    cout << 2;
    return 0;
  }
  long long t = 0, step = 0, m1, m2;
  while (t < a) {
    step++;
    t += po[step];
  }
  long long total = po[step] - 1;
  m1 = (n - total + (t - a)) / po[step - 1];
  if (m1 % 2) m1--;
  t = 0;
  step = 0;
  while (t < a) {
    t += po[step];
    step++;
  }
  total = po[step - 1] - 1;
  m2 = (n - total + (t - a)) / po[step - 1];
  if (!(m2 % 2)) m2--;
  cout << max(m1, m2) << endl;
}
