#include <bits/stdc++.h>
using namespace std;
void pre() {}
long long int n, k;
long long int pru(long long int aa) {
  long long int curr = aa, ret = 0, cu = 1, val = 0;
  while (curr <= n) {
    ret += cu;
    curr *= 2;
    curr += 1;
    cu *= 2;
    val++;
  }
  ret += max((long long int)0, n + 1 - (aa << val));
  return ret;
}
long long int val(long long int oo) {
  long long int bb = 0;
  if (oo % 2 == 0) bb += pru(oo + 1);
  bb += pru(oo);
  return bb;
}
void solve() {
  cin >> n >> k;
  unsigned long long int ans = 0;
  for (long long int i = 62; i >= 0; i--) {
    if (val(ans + ((long long int)1 << i)) >= k) {
      ans += ((long long int)1 << i);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre();
  long long int num = 1;
  for (long long int i = 0; i < num; i++) {
    solve();
  }
}

### Prompt

Please formulate 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;
void pre() {}
long long int n, k;
long long int pru(long long int aa) {
  long long int curr = aa, ret = 0, cu = 1, val = 0;
  while (curr <= n) {
    ret += cu;
    curr *= 2;
    curr += 1;
    cu *= 2;
    val++;
  }
  ret += max((long long int)0, n + 1 - (aa << val));
  return ret;
}
long long int val(long long int oo) {
  long long int bb = 0;
  if (oo % 2 == 0) bb += pru(oo + 1);
  bb += pru(oo);
  return bb;
}
void solve() {
  cin >> n >> k;
  unsigned long long int ans = 0;
  for (long long int i = 62; i >= 0; i--) {
    if (val(ans + ((long long int)1 << i)) >= k) {
      ans += ((long long int)1 << i);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pre();
  long long int num = 1;
  for (long long int i = 0; i < num; i++) {
    solve();
  }
}

```