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;
const int N = 1000 * 1000 + 20;
long long int a[N], m[N];
vector<int> e[N];
long long int n, k;
bool check(long long int x) {
  long long int s = 0, karan;
  if (x % 2 == 1) {
    s = 1;
    karan = 2 * x + 1;
    x = 2 * x;
  } else {
    karan = x + 1;
  }
  while (karan <= n) {
    s += karan - x + 1;
    karan = karan * 2 + 1;
    x *= 2;
  }
  if (x <= n) {
    s += n - x + 1;
  }
  return (k <= s ? true : false);
}
int main() {
  cin >> n >> k;
  if (k == 1) {
    cout << n << "\n";
    return 0;
  }
  long long int l = 1, r = n;
  while ((r - l) > 1) {
    long long int mid = (l + r) / 2;
    if (check(mid + 1)) {
      l = mid + 1;
    } else if (check(mid)) {
      l = mid;
    } else {
      r = mid;
    }
  }
  if (n == 2 and k == 1) {
    cout << 2 << "\n";
    return 0;
  }
  cout << l << "\n";
}
