#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long check(long long x) {
  queue<pair<long long, long long> > q;
  if (1 == x % 2)
    q.push(make_pair(x, x));
  else
    q.push(make_pair(x, x + 1));
  long long ans = 0;
  while (!q.empty()) {
    pair<long long, long long> p = q.front();
    q.pop();
    ans = ans + min(n, p.second) - p.first + 1;
    if (p.first * 2 <= n) q.push(make_pair(p.first * 2, p.second * 2 + 1));
  }
  return ans;
}
int main() {
  scanf("%lld%lld", &n, &k);
  long long l1 = 0, r1 = n;
  while (l1 < r1) {
    long long mid = (l1 + r1 + 1) >> 1;
    if (check(2 * mid) >= k)
      l1 = mid;
    else
      r1 = mid - 1;
  }
  long long l2 = 0, r2 = n;
  while (l2 < r2) {
    long long mid = (l2 + r2 + 1) >> 1;
    if (check(2 * mid + 1) >= k)
      l2 = mid;
    else
      r2 = mid - 1;
  }
  printf("%lld\n", max(2 * l1, 2 * l2 + 1));
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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>
using namespace std;
long long n, k;
long long check(long long x) {
  queue<pair<long long, long long> > q;
  if (1 == x % 2)
    q.push(make_pair(x, x));
  else
    q.push(make_pair(x, x + 1));
  long long ans = 0;
  while (!q.empty()) {
    pair<long long, long long> p = q.front();
    q.pop();
    ans = ans + min(n, p.second) - p.first + 1;
    if (p.first * 2 <= n) q.push(make_pair(p.first * 2, p.second * 2 + 1));
  }
  return ans;
}
int main() {
  scanf("%lld%lld", &n, &k);
  long long l1 = 0, r1 = n;
  while (l1 < r1) {
    long long mid = (l1 + r1 + 1) >> 1;
    if (check(2 * mid) >= k)
      l1 = mid;
    else
      r1 = mid - 1;
  }
  long long l2 = 0, r2 = n;
  while (l2 < r2) {
    long long mid = (l2 + r2 + 1) >> 1;
    if (check(2 * mid + 1) >= k)
      l2 = mid;
    else
      r2 = mid - 1;
  }
  printf("%lld\n", max(2 * l1, 2 * l2 + 1));
  return 0;
}

```