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 T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
long long n, k;
inline long long check(long long x) {
  long long rs = 0, q = 1 + !(x & 1);
  while (x + q <= n) x <<= 1, rs += q, q <<= 1;
  if (n >= x)
    return n - x + 1 + rs;
  else
    return rs;
}
int main() {
  read(n), read(k);
  long long l, r, rs1, rs2, md;
  rs1 = rs2 = 0;
  l = 1, r = n >> 1;
  while (l <= r) {
    md = (l + r) >> 1;
    if (check(md << 1) >= k)
      rs1 = md << 1, l = md + 1;
    else
      r = md - 1;
  }
  l = 0, r = n >> 1;
  while (l <= r) {
    md = (l + r) >> 1;
    if (check(md << 1 | 1) >= k)
      rs2 = md << 1 | 1, l = md + 1;
    else
      r = md - 1;
  }
  return printf("%lld\n", max(rs1, rs2)), 0;
}
