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;
using ilong = int64_t;
using ulong = uint64_t;
ilong N, K;
ilong f(ilong x) {
  if (x > N) return 0;
  int k = 0, dk = (x % 2 == 0);
  ilong ans = 0;
  while ((x << k) <= N) ans += (1ll << (k + dk)), k++;
  return ans - max((x << (k - 1)) + ((1ll << (k - 1 + dk)) - 1) - N, 0ll);
}
ilong bs(ilong lo, ilong hi) {
  while (lo < hi) {
    ilong mi = (hi + lo) / 2;
    if ((mi & 1) != (hi & 1)) mi++;
    if (f(mi) >= K)
      lo = mi;
    else
      hi = mi - 2;
  }
  return lo;
}
void solution() {
  cin >> N >> K;
  if (K == N) {
    cout << 1 << endl;
    return;
  }
  cout << max(bs(1, N - (N % 2 == 0)), bs(2, N - (N % 2 == 1))) << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solution();
}
