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 pow(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  if (x == 0) return 0;
  while (y > 0) {
    if ((y & 1) == 1) res = (res * x) % 1000000007;
    y >>= 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long count(long long v, long long n, long long rc) {
  long long c = 1;
  c = c << (rc);
  c -= 1;
  cerr << v << " " << rc << " -> ";
  v <<= rc;
  if (v <= n) c += (n - v + 1);
  cerr << c << "\n";
  return c;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, k, i;
  cin >> n >> k;
  long long lk = 0, t = k;
  while (t > 0) {
    lk++;
    t >>= 1;
  }
  long long temp = 1;
  temp <<= lk;
  temp--;
  if (k == temp) lk++;
  long long v = n, ck = 1, p = 1, pa = 2, rc = 0;
  while (ck < k) {
    if (v % 2 == 0 && rc == lk - 2) {
      if (v == 2)
        v--;
      else
        v -= (2 - v % 2);
      break;
    }
    if (v % 2 == 1) {
      v--;
      ck += p;
    } else {
      v >>= 1;
      rc++;
      ck++;
      if (v % 2 == 0) ck += count(v + 1, n, rc);
      p += pa;
      pa <<= 1;
    }
    cerr << v << " " << ck << "\n";
  }
  cout << v << "\n";
  return 0;
}
