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;
bool chck(long long x, long long n, long long k) {
  long long j = x;
  long long c = 0;
  while (j <= n) j = j * 2, c++;
  j = j / 2;
  c--;
  long long a;
  if (x % 2 != 0) {
    a = pow(2, c) - 1;
    long long l = pow(2, c);
    a += min(n - j + 1, l);
  } else {
    a = 2 * (pow(2, c) - 1);
    long long l = pow(2, c + 1);
    a += min(n - j + 1, l);
  }
  if (a >= k)
    return true;
  else
    return false;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  long long s = 1, l = n;
  long long ans = 1;
  while (s <= l) {
    long long x = (s + l) / 2;
    long long f1 = 0, f2 = 0;
    if (chck(x, n, k)) {
      f1 = 1;
      ans = max(ans, x);
      s = x + 1;
    }
    if (chck(x + 1, n, k)) {
      f2 = 1;
      ans = max(ans, x + 1);
      s = x + 2;
    }
    if (f1 == 0 && f2 == 0) {
      l = x - 1;
    }
  }
  cout << ans;
}
