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;
long long t = 2;
long long n, k;
long long check(long long x) {
  if (x > n) return 0;
  long long p = 1, cx;
  while (p * x <= n) p *= 2;
  p /= 2;
  cx = p * x;
  long long l = 0, r = p;
  while (r - l > 1) {
    long long m = (l + r) / 2;
    if (cx + m <= n)
      l = m;
    else
      r = m;
  }
  long long ret = 0;
  ret = l + 1;
  while (p) {
    p /= 2;
    ret += p;
  }
  if (x % 2 == 0) ret += check(x + 1);
  return ret;
}
int main() {
  long long x;
  cin >> n >> k;
  long long l = 0, r = n + 1, o = 0, m;
  while (r - l > 1) {
    m = (r + l) / 2;
    if (m != 0) {
      if (check(m + m % 2) >= k && m + m % 2 <= n)
        l = m;
      else
        r = m;
    } else
      r = m;
  }
  long long mx = l + l % 2;
  l = 1;
  r = n + 1;
  while (r - l > 1) {
    m = (r + l) / 2;
    if (check(m + 1 - m % 2) >= k && m + 1 - m % 2 <= n)
      l = m;
    else
      r = m;
  }
  cout << max(mx, l);
}
