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.
#include <bits/stdc++.h>
using namespace std;
long long count(long long a, long long b) {
  long long ans = a % 2 ? 1 : 2;
  long long res = ans;
  while (a * 2 <= b) {
    a *= 2;
    ans *= 2;
    res += ans;
  }
  if (a + ans > b) res -= a + ans - b - 1;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long k, n;
  cin >> n >> k;
  if (k == n) {
    cout << 1;
    return 0;
  }
  if (k == 1) {
    cout << n;
    return 0;
  }
  long long l = 0, r = n + 1;
  while (r - l > 2) {
    long long m = (r - l) / 2 + l;
    if (!(m % 2)) m++;
    auto t = count(m, n);
    if (t < k)
      r = m;
    else
      l = m;
  }
  vector<long long> ans;
  for (long long i = l; i <= r; i++) {
    ans.push_back(count(i, n));
  }
  long long res = -1;
  long long pos = 0;
  for (long long i = l; i <= r; i++) {
    if (ans[i - l] >= res && ans[i - l] >= k) {
      pos = i;
      res = max(res, ans[i - l]);
    }
  }
  l = 0;
  r = n + 1;
  while (r - l > 2) {
    long long m = (r - l) / 2 + l;
    if ((m % 2)) m++;
    auto t = count(m, n);
    if (t < k)
      r = m;
    else
      l = m;
  }
  ans.clear();
  for (long long i = l; i <= r; i++) {
    ans.push_back(count(i, n));
  }
  res = -1;
  long long pos_ = 0;
  for (long long i = l; i <= r; i++) {
    if (ans[i - l] >= res && ans[i - l] >= k) {
      pos_ = i;
      res = max(res, ans[i - l]);
    }
  }
  cout << max(pos, pos_);
  return 0;
}
