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 n, k;
inline long long check(long long x) {
  queue<pair<long long, long long> > p;
  if (x & 1)
    p.push(make_pair(x, x));
  else
    p.push(make_pair(x, x + 1));
  long long ans = 0;
  while (p.size()) {
    pair<long long, long long> zz = p.front();
    p.pop();
    ans += min(zz.second, n) - zz.first + 1;
    if ((zz.first << 1) <= n)
      p.push(make_pair(zz.first << 1, zz.second << 1 | 1));
  }
  return ans;
}
int main() {
  scanf("%lld%lld", &n, &k);
  long long l = 1, r = n / 2ll;
  long long ans = 0;
  while (l <= r) {
    long long mid = l + r >> 1ll;
    long long cnt = check(mid << 1);
    if (cnt >= k) {
      ans = max(ans, mid << 1);
      l = mid + 1ll;
    } else
      r = mid - 1;
  }
  l = 0, r = n / 2ll;
  while (l <= r) {
    long long mid = l + r >> 1ll;
    long long cnt = check(mid << 1 | 1);
    if (cnt >= k) {
      ans = max(ans, mid << 1 | 1);
      l = mid + 1;
    } else
      r = mid - 1;
  }
  cout << ans << endl;
}
