#include <bits/stdc++.h>
using namespace std;
int const N = 2005;
long long f(long long x, long long n) {
  long long t = (x % 2 == 1 ? 1 : 2);
  long long h = 1;
  long long res = 0;
  while (true) {
    if ((x * h) > n) return res;
    res += (t - (x * h + t - 1 - min(x * h + t - 1, n)));
    t <<= 1;
    h <<= 1;
  }
  return res;
}
long long bs(long long t, long long n, long long k) {
  long long l = (t ? t : 2), r = (n % 2 == t ? n : n - 1);
  long long res = 0;
  while (l <= r) {
    long long md = (l + r) / 2;
    md = (md % 2 == t ? md : md - 1);
    long long x = f(md, n);
    if (x >= k) {
      res = max(res, md);
      l = md + 2;
    } else
      r = md - 2;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  cout << max(bs(0, n, k), bs(1, n, k)) << endl;
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int const N = 2005;
long long f(long long x, long long n) {
  long long t = (x % 2 == 1 ? 1 : 2);
  long long h = 1;
  long long res = 0;
  while (true) {
    if ((x * h) > n) return res;
    res += (t - (x * h + t - 1 - min(x * h + t - 1, n)));
    t <<= 1;
    h <<= 1;
  }
  return res;
}
long long bs(long long t, long long n, long long k) {
  long long l = (t ? t : 2), r = (n % 2 == t ? n : n - 1);
  long long res = 0;
  while (l <= r) {
    long long md = (l + r) / 2;
    md = (md % 2 == t ? md : md - 1);
    long long x = f(md, n);
    if (x >= k) {
      res = max(res, md);
      l = md + 2;
    } else
      r = md - 2;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  cout << max(bs(0, n, k), bs(1, n, k)) << endl;
  return 0;
}

```