#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
long long n, k;
long long calc(long long x) {
  long long y = 0, tmp = n;
  int diff = 0;
  while ((int)log2(x) < (int)log2(tmp)) {
    if (tmp & 1) {
      y |= ((long long)1 << diff);
    }
    tmp >>= 1;
    diff++;
  }
  if (tmp > x) {
    return ((long long)1 << (diff + 1)) - 1;
  } else if (tmp < x) {
    return ((long long)1 << diff) - 1;
  } else {
    return ((long long)1 << diff) + y;
  }
}
bool valid(long long val) { return calc(val) + calc(val + 1) >= k; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  long long l = 0, r = n / 2;
  while (l < r) {
    long long mid = l + (r - l + 1) / 2;
    if (valid(2 * mid)) {
      l = mid;
    } else {
      r = mid - 1;
    }
  }
  long long ans = 2 * l;
  cout << (calc(ans + 1) >= k ? ans + 1 : ans) << '\n';
}

### Prompt

Please formulate 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;
const int INF = 1e9;
long long n, k;
long long calc(long long x) {
  long long y = 0, tmp = n;
  int diff = 0;
  while ((int)log2(x) < (int)log2(tmp)) {
    if (tmp & 1) {
      y |= ((long long)1 << diff);
    }
    tmp >>= 1;
    diff++;
  }
  if (tmp > x) {
    return ((long long)1 << (diff + 1)) - 1;
  } else if (tmp < x) {
    return ((long long)1 << diff) - 1;
  } else {
    return ((long long)1 << diff) + y;
  }
}
bool valid(long long val) { return calc(val) + calc(val + 1) >= k; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  long long l = 0, r = n / 2;
  while (l < r) {
    long long mid = l + (r - l + 1) / 2;
    if (valid(2 * mid)) {
      l = mid;
    } else {
      r = mid - 1;
    }
  }
  long long ans = 2 * l;
  cout << (calc(ans + 1) >= k ? ans + 1 : ans) << '\n';
}

```