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>
#pragma GCC optimize("-O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long MAXN = 1123456;
const long long N = 2e6;
const long long inf = 1e18;
mt19937_64 rnd(chrono::system_clock::now().time_since_epoch().count());
template <typename T>
void vout(T s) {
  cout << s << endl;
  exit(0);
}
long long n, k;
long long f(long long x) {
  if (x > n) return 0;
  if (x & 1) return 1 + f(x << 1);
  long long ans = 0, p = 3;
  for (long long st = 0, step = 1;; st++, step *= 2) {
    if (st == 0) {
      long long l = x * step, r = x * step + (step * 2) - 1;
      if (l > n) break;
      ans += min(r, n) - l + 1;
      if (r >= n) break;
    } else if (st == 1) {
      long long l = x * step, r = x * step + (step * 2) - 1;
      if (l > n) break;
      ans += min(r, n) - l + 1;
      if (r >= n) break;
    } else {
      p *= 2;
      long long l = x * step, r = x * step + (step * 2) - 1;
      if (l > n) break;
      ans += min(r, n) - l + 1;
      if (r >= n) break;
    }
  }
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  long long l = 0, r = n / 2;
  long long ans = 0;
  while (l < r) {
    long long c = (l + r + 1) >> 1;
    long long val = c * 2;
    if (f(val) < k)
      r = c - 1;
    else
      l = c;
  }
  ans = l * 2;
  l = 0, r = (n + 1) / 2;
  while (l < r) {
    long long c = (l + r + 1) >> 1;
    long long val = c * 2 - 1;
    if (f(val) < k)
      r = c - 1;
    else
      l = c;
  }
  ans = max(ans, (2 * l) - 1);
  cout << ans << "\n";
  return 0;
}
