#include <bits/stdc++.h>
using namespace std;
using lli = long long int;
lli n, k;
lli get_count(lli x, lli p, lli q) {
  lli lo = p * x, hi = p * x + q - 1;
  if (hi > n) return max(n - lo + 1, 0LL);
  return q + get_count(x, p << 1, q << 1);
}
lli even(lli x) { return x - (x & 1); }
lli odd(lli x) {
  if (x & 1) return x;
  return x - 1;
}
void solve() {
  cin >> n >> k;
  lli lo, hi, mid, ans = 0;
  lo = 2, hi = even(n);
  while (lo <= hi) {
    mid = even((lo + hi) / 2);
    if (get_count(mid, 1, 2) >= k) {
      ans = max(ans, mid);
      lo = mid + 2;
    } else
      hi = mid - 2;
  }
  lo = 1, hi = odd(n);
  while (lo <= hi) {
    mid = odd((lo + hi) / 2);
    if (get_count(mid, 1, 1) >= k) {
      ans = max(ans, mid);
      lo = mid + 2;
    } else
      hi = mid - 2;
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### Prompt

In CPP, your task is to solve 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;
using lli = long long int;
lli n, k;
lli get_count(lli x, lli p, lli q) {
  lli lo = p * x, hi = p * x + q - 1;
  if (hi > n) return max(n - lo + 1, 0LL);
  return q + get_count(x, p << 1, q << 1);
}
lli even(lli x) { return x - (x & 1); }
lli odd(lli x) {
  if (x & 1) return x;
  return x - 1;
}
void solve() {
  cin >> n >> k;
  lli lo, hi, mid, ans = 0;
  lo = 2, hi = even(n);
  while (lo <= hi) {
    mid = even((lo + hi) / 2);
    if (get_count(mid, 1, 2) >= k) {
      ans = max(ans, mid);
      lo = mid + 2;
    } else
      hi = mid - 2;
  }
  lo = 1, hi = odd(n);
  while (lo <= hi) {
    mid = odd((lo + hi) / 2);
    if (get_count(mid, 1, 1) >= k) {
      ans = max(ans, mid);
      lo = mid + 2;
    } else
      hi = mid - 2;
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```