#include <bits/stdc++.h>
const long long MOD = 1e9 + 7;
using pii = std::pair<long long, long long>;
using namespace std;
const long long maxn = 1e6 + 5;
long long n, k, cnt[maxn];
long long check(long long x) {
  long long totalcnt = 0, curlength = 1, prevend = -1, curstart, curend;
  if (x % 2 == 0) curlength++;
  while (x <= n) {
    curstart = max(x, prevend + 1), curend = min(x + curlength - 1, n);
    if (curstart <= curend) {
      prevend = curend;
      totalcnt += (curend - curstart + 1);
    }
    x *= 2;
    curlength *= 2;
  }
  return totalcnt >= k;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  long long lo = 0, hi = n / 2, mid;
  while (lo < hi) {
    mid = (lo + hi + 1) / 2;
    if (check(2 * mid + 1))
      lo = mid;
    else
      hi = mid - 1;
  }
  assert(lo == hi);
  long long oddans = 2 * lo + 1;
  lo = 0, hi = n / 2, mid;
  while (lo < hi) {
    mid = (lo + hi + 1) / 2;
    if (!mid || check(2 * mid))
      lo = mid;
    else
      hi = mid - 1;
  }
  assert(lo == hi);
  long long evenans = 2 * lo;
  cout << max(oddans, evenans) << "\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 long long MOD = 1e9 + 7;
using pii = std::pair<long long, long long>;
using namespace std;
const long long maxn = 1e6 + 5;
long long n, k, cnt[maxn];
long long check(long long x) {
  long long totalcnt = 0, curlength = 1, prevend = -1, curstart, curend;
  if (x % 2 == 0) curlength++;
  while (x <= n) {
    curstart = max(x, prevend + 1), curend = min(x + curlength - 1, n);
    if (curstart <= curend) {
      prevend = curend;
      totalcnt += (curend - curstart + 1);
    }
    x *= 2;
    curlength *= 2;
  }
  return totalcnt >= k;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  long long lo = 0, hi = n / 2, mid;
  while (lo < hi) {
    mid = (lo + hi + 1) / 2;
    if (check(2 * mid + 1))
      lo = mid;
    else
      hi = mid - 1;
  }
  assert(lo == hi);
  long long oddans = 2 * lo + 1;
  lo = 0, hi = n / 2, mid;
  while (lo < hi) {
    mid = (lo + hi + 1) / 2;
    if (!mid || check(2 * mid))
      lo = mid;
    else
      hi = mid - 1;
  }
  assert(lo == hi);
  long long evenans = 2 * lo;
  cout << max(oddans, evenans) << "\n";
  return 0;
}

```