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 int n;
int to_bin(long long int x) {
  int res = 0;
  while (x) {
    res++;
    x /= 2;
  }
  return res;
}
long long int f(long long int x) {
  int l1 = to_bin(n);
  int l2 = to_bin(x);
  int s = l1 - l2;
  long long int d = n / (1ll << s);
  long long int ans = 0;
  if (d == x) {
    ans = n % (1ll << s);
    ans += (1ll << (s));
  } else if (d > x) {
    ans = (1ll << (s + 1)) - 1;
  } else {
    ans = (1ll << (s)) - 1;
  }
  if (!(x & 1)) ans += f(x + 1);
  return ans;
}
int main() {
  cin >> n;
  long long int k;
  cin >> k;
  long long int l = 2;
  long long int h = n - n % 2;
  long long int ans = 1;
  while (l <= h) {
    long long int mid = (l + h) >> 1;
    if (mid & 1) mid -= 1;
    if (f(mid) >= k) {
      ans = mid;
      l = mid + 2;
    } else {
      h = mid - 2;
    }
  }
  if (f(ans + 1) >= k) {
    ans++;
  }
  cout << ans << "\n";
  return 0;
}
