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;
function<void(void)> ____ = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
using LL = int_fast64_t;
LL n, k;
bool check(LL mid, bool tag) {
  LL tot = 0;
  int p = 0;
  while (mid * (1ll << (p + 1)) <= n) p++;
  for (int i = 0; i < p; i++) tot += (1ll << i);
  tot += min(1ll << p, n - mid * (1ll << p) + 1);
  if (tag) tot--;
  return tot >= k;
}
LL gaoodd() {
  LL pl = 1, pr = n & 1 ? n : n - 1;
  LL l = (pl + 1) >> 1, r = (pr + 1) >> 1;
  while (l <= r) {
    LL mid = (l + r) >> 1;
    if (check(mid * 2ll - 1, false))
      l = mid + 1;
    else
      r = mid - 1;
  }
  return (r << 1) - 1;
}
LL gaoeven() {
  LL pl = 2, pr = n & 1 ? n - 1 : n;
  LL l = pl >> 1, r = pr >> 1;
  if (pl > pr) return 0;
  while (l <= r) {
    LL mid = (l + r) >> 1;
    if (check(mid, true))
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r << 1;
}
int main() {
  cin >> n >> k;
  cout << max(gaoodd(), gaoeven()) << endl;
  return 0;
}
