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 denom = 1e9 + 7;
long long n, k;
unordered_map<long long, long long> mp;
long long calc(long long m) {
  if (m == 1) return n;
  if (m > n) return 0;
  if (m == n) return 1;
  long long res = 0;
  if (m % 2 == 0) {
    long long p = 1;
    long long a = 2;
    res = 2;
    while (true) {
      m *= 2;
      if (m > n) break;
      a *= 2;
      p++;
      res += min(n, m + a - 1) - m + 1;
    }
  } else {
    res = 1 + calc(2 * m);
  }
  return res;
}
bool fit(long long m) { return calc(m) >= k; }
int main(int argc, const char* argv[]) {
  cin >> n >> k;
  long long l = 1;
  long long h = n;
  while (l + 1 < h) {
    long long mid = l + (h - l) / 2;
    if (fit(mid + 1)) {
      l = mid + 1;
    } else {
      if (fit(mid)) {
        l = mid;
      } else {
        h = mid - 1;
      }
    }
  }
  cout << (fit(h) ? h : l);
  return 0;
}
