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;
int main() {
  long long(n), (k);
  scanf("%lld%lld", &(n), &(k));
  ;
  long long n2 = n;
  vector<long long> vec;
  while (n) {
    vec.push_back(n);
    if (n - 1 > 0) vec.push_back(n - 1);
    n /= 2;
  }
  long long maxi = 1;
  auto allok = [&](long long v) {
    long long mi = v, ma = (v % 2) ? v : v + 1;
    long long c = ma - mi + 1;
    while (ma <= n2) {
      mi *= 2;
      ma = 2 * ma + 1;
      if (mi < n2) c += min(ma, n2) - mi + 1;
    }
    return c >= k;
  };
  for (auto i : vec) {
    while (i <= n2) {
      if (allok(i)) maxi = max(maxi, i);
      i *= 2;
    }
  }
  printf("%lld\n", maxi);
  return 0;
}
