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;
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class L, class R, class M>
ostream& operator<<(ostream& os, tuple<L, R, M> P) {
  return os << "(" << get<0>(P) << "," << get<1>(P) << "," << get<2>(P) << ")";
}
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "{ ";
  for (auto v : V) os << v << " ";
  return os << "}";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long maxn = (long long)1e5 + 5;
const long double EPS = 1e-9;
const long long INF = (long long)1e9 + 10;
const long long mod = (long long)1e9 + 7;
long long n, k;
long long get(long long L, long long R) {
  if (R > n) {
    R = n;
    return (L > n ? 0LL : R - L + 1);
  }
  long long ret = R - L + 1;
  ret += get(2 * L, 2 * R + 1);
  return ret;
}
long long get_paths(long long number) {
  if (number % 2 == 0) {
    return get(number, number + 1);
  } else {
    return get(number, number);
  }
  assert(false);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  long long answer = 0, bits = 61;
  for (long long bit = bits - 1; bit >= 0; bit--) {
    if (get_paths(answer | (1LL << bit)) >= k) {
      answer |= (1LL << bit);
    }
  }
  cout << answer << '\n';
  return 0;
}
