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;
const int maxn = 1e5 + 10;
long long n, k;
int main() {
  cin >> n >> k;
  long long now = n, cnt = 1, a = 1, b = (n & 1) ? 2 : 1, c = (n & 1) ? 1 : 2,
            d = (n & 1) ? 2 : 1;
  if (n == k) return puts("1"), 0;
  while (1) {
    if (a >= k) {
      cout << now << '\n';
      return 0;
    }
    if (b >= k) {
      cout << now - 1ll << '\n';
      return 0;
    }
    if (c >= k) {
      cout << now - 2ll << '\n';
      return 0;
    }
    if (d >= k) {
      cout << now - 3ll << '\n';
      return 0;
    }
    if (now & 1ll) {
      now >>= 1ll;
      if (now & 1ll) {
        a = b + 1ll;
      } else {
        a = b + (1ll << cnt);
      }
    } else {
      now >>= 1ll;
      if (now & 1ll) {
        a++;
      } else {
        a = a + (1ll << cnt);
      }
    }
    cnt++;
    b = (now & 1ll) ? a + ((1ll << cnt) - 1ll) : ((1ll << cnt) - 1ll);
    c = (now & 1ll) ? ((1ll << cnt) - 1ll) : ((1ll << cnt) - 1ll) << 1ll;
    if (now == 6)
      d = a + 1;
    else if (now == 5)
      d = b + c + 1;
    else if (now == 4)
      d = n;
    else {
      d = (now & 1ll) ? c + 1 + ((1ll << (cnt - 1)) - 1ll) * 2
                      : (1ll << cnt) - 1ll;
    }
  }
}
