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 <typename T1, typename T2>
inline void mine(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void maxe(T1 &x, T2 y) {
  if (x < y) x = y;
}
ostream &operator<<(ostream &a, const vector<long long> &b) {
  for (auto k : b) cout << k << ' ';
  return a;
}
const long long MOD = 1000000007;
const long long INF = 1000000050;
const long long BIG = (long long)2e18 + 50;
const long long MX = 200010;
const double EPS = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long n, k;
long long f(long long x) {
  if (x > n) return 0;
  long long res = 0;
  if (x % 2 == 0) res = f(x + 1);
  long long len = 0;
  while ((x << (len + 1)) <= n) {
    res += 1ll << len;
    ++len;
  }
  long long delta = n + 1 - (x << len);
  delta = min(delta, 1ll << len);
  return (res + delta);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  long long l = 0;
  long long r = n / 2 + 3;
  while (r - l > 1) {
    long long m = (r + l) / 2;
    if (f(2 * m + 1) >= k)
      l = m;
    else
      r = m;
  }
  long long res1 = 2 * l + 1;
  l = 1;
  r = n / 2 + 3;
  while (r - l > 1) {
    long long m = (r + l) / 2;
    if (f(2 * m) >= k)
      l = m;
    else
      r = m;
  }
  long long res2 = 2 * l;
  if (f(res2) < k) res2 = -INF;
  if (f(res1) < k) res1 = -INF;
  cout << max(res1, res2) << '\n';
  assert(res2 != -INF || res1 != -INF);
}
