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;
void prb(int upper) {
  for (int i = 10; i >= 0; i--) {
    if ((upper & (1 << i)) > 0)
      cout << 1;
    else
      cout << 0;
  }
  cout << '\n';
}
long long n, k;
long long cnt(long long K) {
  long long lower = K;
  long long upper = K + (K % 2 == 0 ? 1 : 0);
  long long tot = 0;
  while (upper <= n) {
    tot += upper - lower + 1;
    lower = lower << 1;
    upper = (upper << 1) + 1;
  }
  upper = min(upper, n);
  tot += max(upper - lower + 1, 0LL);
  return tot;
}
long long solveeven() {
  long long r = n;
  if (r & 1 != 0) r--;
  long long l = 1;
  r = r / 2;
  while (l < r) {
    long long m = l + (r - l + 1) / 2;
    m = m * 2;
    if (cnt(m) >= k) {
      l = m / 2;
    } else {
      r = m / 2 - 1;
    }
  }
  if (cnt(2 * l) < k) return 0;
  return l * 2;
}
long long solveodd() {
  long long r = n;
  if (r & 1 == 0) r--;
  long long l = 0;
  r = r / 2;
  while (l < r) {
    long long m = (l) + (r - l + 1) / 2;
    m = m * 2 + 1;
    if (cnt(m) >= k) {
      l = m / 2;
    } else {
      r = m / 2 - 1;
    }
  }
  return l * 2 + 1;
}
int main() {
  cin >> n >> k;
  long long ans = 0;
  ans = max(solveeven(), solveodd());
  cout << ans << '\n';
}
