#include <bits/stdc++.h>
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long n, k;
bool check(long long x) {
  long long tmp = x;
  long long cnt = 0;
  long long cur;
  if (x & 1)
    cur = 1;
  else
    cur = 2;
  while (x <= n) {
    cnt += min(cur, n - x + 1);
    x <<= 1;
    cur <<= 1;
  }
  if (cnt >= k) return true;
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  long long l = 0, r = (n - 1) / 2;
  long long ans = 1;
  while (l < r) {
    long long mid = (l + r + 1) >> 1;
    long long x = mid * 2 + 1;
    if (check(x))
      l = mid;
    else
      r = mid - 1;
  }
  ans = max(ans, 2 * l + 1);
  l = 0, r = n / 2;
  while (l < r) {
    long long mid = (l + r + 1) >> 1;
    long long x = mid * 2;
    if (check(x))
      l = mid;
    else
      r = mid - 1;
  }
  ans = max(ans, 2 * l);
  cout << ans << '\n';
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long n, k;
bool check(long long x) {
  long long tmp = x;
  long long cnt = 0;
  long long cur;
  if (x & 1)
    cur = 1;
  else
    cur = 2;
  while (x <= n) {
    cnt += min(cur, n - x + 1);
    x <<= 1;
    cur <<= 1;
  }
  if (cnt >= k) return true;
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  long long l = 0, r = (n - 1) / 2;
  long long ans = 1;
  while (l < r) {
    long long mid = (l + r + 1) >> 1;
    long long x = mid * 2 + 1;
    if (check(x))
      l = mid;
    else
      r = mid - 1;
  }
  ans = max(ans, 2 * l + 1);
  l = 0, r = n / 2;
  while (l < r) {
    long long mid = (l + r + 1) >> 1;
    long long x = mid * 2;
    if (check(x))
      l = mid;
    else
      r = mid - 1;
  }
  ans = max(ans, 2 * l);
  cout << ans << '\n';
  return 0;
}

```