#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
#pragma GCC target("avx2")
using namespace std;
long long count0(long long n, long long x) {
  if (x > n) return 0;
  long long cnt = 0, bin = 1;
  while (bin * x <= n) {
    if (bin * 2 + x * bin - 1 >= n) {
      cnt += n - x * bin + 1;
      return cnt;
    } else {
      cnt += bin * 2;
    }
    bin *= 2;
  }
  return cnt;
}
long long count1(long long n, long long x) {
  if (x > n) return 0;
  long long cnt = 0, bin = 1;
  while (bin * x <= n) {
    if (bin + x * bin - 1 >= n) {
      cnt += n - x * bin + 1;
      return cnt;
    } else {
      cnt += bin;
    }
    bin *= 2;
  }
  return cnt;
}
signed main() {
  long long n, k;
  cin >> n >> k;
  long long l = 0, r = n + 1, res = 0;
  while (r > l + 1) {
    long long mid = (r + l) / 2;
    if (count0(n, 2 * mid) >= k) {
      res = max(res, 2 * mid);
      l = mid;
    } else
      r = mid;
  }
  long long ll = 0, rr = n + 1;
  while (rr > ll + 1) {
    long long mid = (rr + ll) / 2;
    if (count1(n, 2 * mid - 1) >= k) {
      res = max(res, 2 * mid - 1);
      ll = mid;
    } else
      rr = mid;
  }
  cout << res << " ";
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#pragma GCC optimize("-O3")
#pragma GCC target("avx2")
using namespace std;
long long count0(long long n, long long x) {
  if (x > n) return 0;
  long long cnt = 0, bin = 1;
  while (bin * x <= n) {
    if (bin * 2 + x * bin - 1 >= n) {
      cnt += n - x * bin + 1;
      return cnt;
    } else {
      cnt += bin * 2;
    }
    bin *= 2;
  }
  return cnt;
}
long long count1(long long n, long long x) {
  if (x > n) return 0;
  long long cnt = 0, bin = 1;
  while (bin * x <= n) {
    if (bin + x * bin - 1 >= n) {
      cnt += n - x * bin + 1;
      return cnt;
    } else {
      cnt += bin;
    }
    bin *= 2;
  }
  return cnt;
}
signed main() {
  long long n, k;
  cin >> n >> k;
  long long l = 0, r = n + 1, res = 0;
  while (r > l + 1) {
    long long mid = (r + l) / 2;
    if (count0(n, 2 * mid) >= k) {
      res = max(res, 2 * mid);
      l = mid;
    } else
      r = mid;
  }
  long long ll = 0, rr = n + 1;
  while (rr > ll + 1) {
    long long mid = (rr + ll) / 2;
    if (count1(n, 2 * mid - 1) >= k) {
      res = max(res, 2 * mid - 1);
      ll = mid;
    } else
      rr = mid;
  }
  cout << res << " ";
}

```