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;
const long long maxn = 1e18 + 1000;
const long long alpha = 29;
const int inf = 1e9;
const int blsz = 800;
const long long mod = 998244353;
void init() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
long long count(long long m, long long n) {
  if (m > n) return 0;
  long long ans = 1;
  if (m % 2 == 0) ans += count(m + 1, n);
  for (long long i = 1;; i++) {
    long long last = (m << i) + (1LL << (i)) - 1;
    if (last <= n)
      ans += (1LL << (i));
    else {
      ans += max(0LL, n - (m << i) + 1);
      break;
    }
  }
  return ans;
}
int main() {
  init();
  long long n, k;
  cin >> n >> k;
  long long l = 1;
  long long r = maxn / 2;
  while (l < r) {
    long long med = (l + r + 1) / 2;
    if (count(2 * med, n) < k)
      r = med - 1;
    else
      l = med;
  }
  long long a1 = 2 * l;
  l = 0;
  r = maxn / 2;
  while (l < r) {
    long long med = (l + r + 1) / 2;
    if (count(2 * med + 1, n) < k)
      r = med - 1;
    else
      l = med;
  }
  long long a2 = 2 * l + 1;
  if (count(a1, n) < k)
    cout << a2;
  else if (count(a2, n) < k)
    cout << a1;
  else
    cout << max(a1, 2 * l + 1);
  return 0;
}
