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;
long long ocheck(long long num, long long n, long long k) {
  long long ans = 1;
  long long curr = 2 * num;
  long long mul = 2;
  while (num * mul + mul - 1 <= n) {
    long long add = min(2 * mul * num - curr, mul);
    ans += add;
    mul *= 2;
    curr = num * mul;
  }
  long long tt = 0;
  long long add = max(n - curr + 1, tt);
  ans += add;
  return ans;
}
long long echeck(long long num, long long n, long long k) {
  long long ans = 1;
  if (num + 1 <= n) ans++;
  long long curr = 2 * num;
  long long mul = 2;
  while (num * mul + 2 * mul - 1 <= n) {
    long long add = min(2 * mul * num - curr, 2 * mul);
    ans += add;
    mul *= 2;
    curr = num * mul;
  }
  long long tt = 0;
  long long add = max(n - curr + 1, tt);
  ans += add;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long n, k;
    cin >> n >> k;
    long long min = 1;
    long long maxi = n;
    if (n % 2 == 0) maxi--;
    long long mid = (min + maxi) / 2;
    while (min <= maxi) {
      mid = (min + maxi) / 2;
      if (mid % 2 == 0) mid--;
      mid = max(mid, min);
      long long ans = 0;
      ans = ocheck(mid, n, k);
      if (ans >= k)
        min = mid + 2;
      else
        maxi = mid - 2;
    }
    long long one = min - 2;
    min = 2;
    maxi = n;
    if (n % 2 != 0) maxi--;
    mid = (min + maxi) / 2;
    while (min <= maxi) {
      mid = (min + maxi) / 2;
      if (mid % 2 != 0) mid--;
      mid = max(mid, min);
      long long ans = 0;
      ans = echeck(mid, n, k);
      if (ans >= k)
        min = mid + 2;
      else
        maxi = mid - 2;
    }
    long long two = min - 2;
    cout << max(one, two);
  }
  return 0;
}
