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;
const int N = 2e5 + 10;
long long n, k;
bool ok(long long x) {
  long long cnt = 0, pw = 2;
  if (x & 1) cnt++, x <<= 1;
  while (x <= n) {
    cnt += min(n - x + 1, pw);
    x *= 2;
    pw *= 2;
  }
  return (cnt >= k);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  if (k == 1)
    return cout << n << "\n", 0;
  else if (k == 2)
    return cout << n - 1 - ((n & 1) ^ 1) << "\n", 0;
  long long low = 0, high = n / 2 + 1;
  while (high - low > 1) {
    long long mid = (low + high) >> 1;
    if (ok(mid * 2))
      low = mid;
    else
      high = mid;
  }
  long long ans = low * 2 + ok(low * 2 | 1);
  cout << ans << "\n";
}
