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;
vector<pair<pair<long long, long long>, pair<long long, long long> > > lol;
long long get(long long m, long long n) {
  if (m > n) return 0;
  if (m == n) return 1;
  if (m % 2) {
    for (long long i = 0; i < lol.size(); ++i)
      if (lol[i].first.first <= m && lol[i].first.second >= m)
        return lol[i].second.first;
  }
  for (long long i = 0; i < lol.size(); ++i)
    if (lol[i].first.first <= m && lol[i].first.second >= m)
      return lol[i].second.second;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long re = n, ev = 2, od = 1;
  lol.push_back({{re, re}, {1, 1}});
  while (re > 0) {
    lol.push_back({{re / 2 + 1, re - 1}, {od, ev}});
    re = re / 2;
    lol.push_back(
        {{re, re}, {get(2 * re, n) + 1, get(2 * re, n) + get(re + 1, n) + 1}});
    if (re % 2) {
      --re;
      lol.push_back(
          {{re, re},
           {get(2 * re, n) + 1, get(2 * re, n) + get(re + 1, n) + 1}});
    }
    long long tod = od;
    od = ev + 1;
    ev = (od + ev + 1);
  }
  for (long long i = 0; i < lol.size(); ++i) {
  }
  long long l = 0, r = n + 1;
  while (l < r - 1) {
    long long m = (l + r) / 2;
    if (get(2 * m, n) >= k)
      l = m;
    else
      r = m;
  }
  long long l1 = 0, r1 = n + 1;
  while (l1 < r1 - 1) {
    long long m = (l1 + r1) / 2;
    if (get(2 * m + 1, n) >= k)
      l1 = m;
    else
      r1 = m;
  }
  cout << max(2 * l, 2 * l1 + 1);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
