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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k, ans = 1, p, temp;
  cin >> n >> k;
  if (n == k) {
    cout << 1;
    return 0;
  }
  if (k == 1) {
    cout << n;
    return 0;
  }
  long long s, e, m;
  s = 1, e = (n - 1) / 2;
  while (s <= e) {
    m = (s + e) / 2;
    p = log2((n + 1) / (2 * m + 2)) + 1;
    temp = 2 * ((1LL << p) - 1) + max(1 + n - (1LL << p) * 2 * m, 0LL);
    if (temp >= k) {
      s = m + 1;
      ans = max(ans, 2 * m);
    } else
      e = m - 1;
  }
  s = 0, e = (n - 3) / 4;
  while (s <= e) {
    m = (s + e) / 2;
    p = log2((n + 1) / (4 * m + 4)) + 1;
    temp = 1 + 2 * ((1LL << p) - 1) +
           max(1 + n - (1LL << (p + 1)) * (2 * m + 1), 0LL);
    if (temp >= k) {
      s = m + 1;
      ans = max(ans, 2 * m + 1);
    } else
      e = m - 1;
  }
  cout << ans;
  return 0;
}
