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 ll = long long;
using namespace std;
namespace Debug {
template <class A, class B>
ostream &operator<<(ostream &out, pair<A, B> &p) {
  out << "(" << p.first << ", " << p.second << ")";
  return out;
}
template <class T>
ostream &operator<<(ostream &out, vector<T> &v) {
  out << "{";
  string sep;
  for (T el : v) out << sep << el, sep = ", ";
  out << "}";
  return out;
}
void debug_out() { cerr << endl; }
template <class Head, class... Tail>
void debug_out(Head head, Tail... tail) {
  cerr << ' ' << head;
  debug_out(tail...);
}
}  // namespace Debug
using namespace Debug;
ll get(ll n, ll val) {
  if (val > n) return 0;
  ll ans = 1 + (val + 1 <= n && val % 2 == 0);
  ;
  for (ll k = 1; (val << k) <= n; k++) {
    ll cur_val = (val << k);
    ll cnt = (1LL << k);
    ans += min(cnt, n - cur_val + 1);
    ;
    if (val % 2 == 0) ans += min(cnt, max(0LL, n - cur_val - (1LL << k) + 1));
    ;
  };
  return ans;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  get(20, 2);
  ll n, k;
  cin >> n >> k;
  ll L = 0, R = (n + 1) / 2 + 1;
  while (R - L > 1) {
    ll mid = (L + R) / 2;
    if (get(n, mid * 2 + 1) >= k)
      L = mid;
    else
      R = mid;
  }
  ll ans = 2 * L + 1;
  L = 0, R = (n + 1) / 2 + 1;
  while (R - L > 1) {
    ll mid = (L + R) / 2;
    if (get(n, mid * 2) >= k)
      L = mid;
    else
      R = mid;
  }
  ans = max(ans, 2 * L);
  cout << ans << '\n';
}
