#include <bits/stdc++.h>
using namespace std;
long long count(long long s, long long l) {
  long long i = 1, ans = 0;
  if (s % 2 == 0) {
    i = 2;
  }
  while (s + i - 1 <= l) {
    ans += i;
    i *= 2;
    s *= 2;
  }
  long long t = max(0ll, l - s + 1);
  return ans + t;
}
int main() {
  long long n, k, j, h = 0, l, p, b = 1, a = 1, mid, i;
  cin >> n >> k;
  i = 1, j = n;
  while (i <= j) {
    mid = (i + j) / 2;
    if (mid % 2 != 0) mid++;
    if (count(mid, n) >= k) {
      a = mid;
      i = mid + 2;
    } else
      j = mid - 2;
    h++;
    if (h > 1000) break;
  }
  i = 1, j = n;
  while (i <= j) {
    mid = (i + j) / 2;
    if (mid % 2 == 0) mid++;
    if (count(mid, n) >= k) {
      b = mid;
      i = mid + 2;
    } else
      j = mid - 2;
    h++;
    if (h > 2000) break;
  }
  cout << max(a, b);
}

### 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;
long long count(long long s, long long l) {
  long long i = 1, ans = 0;
  if (s % 2 == 0) {
    i = 2;
  }
  while (s + i - 1 <= l) {
    ans += i;
    i *= 2;
    s *= 2;
  }
  long long t = max(0ll, l - s + 1);
  return ans + t;
}
int main() {
  long long n, k, j, h = 0, l, p, b = 1, a = 1, mid, i;
  cin >> n >> k;
  i = 1, j = n;
  while (i <= j) {
    mid = (i + j) / 2;
    if (mid % 2 != 0) mid++;
    if (count(mid, n) >= k) {
      a = mid;
      i = mid + 2;
    } else
      j = mid - 2;
    h++;
    if (h > 1000) break;
  }
  i = 1, j = n;
  while (i <= j) {
    mid = (i + j) / 2;
    if (mid % 2 == 0) mid++;
    if (count(mid, n) >= k) {
      b = mid;
      i = mid + 2;
    } else
      j = mid - 2;
    h++;
    if (h > 2000) break;
  }
  cout << max(a, b);
}

```