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;
void pr_init() {}
long long n, k;
long long raise(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 != 0) res = res * a;
    b /= 2;
    a = a * a;
  }
  return res;
}
long long odd(long long x) {
  long long cn = 0;
  for (long long i = 1; i <= 64; i++) {
    if (raise(2, i) * (x + 1) - 1 <= n) {
      cn += raise(2, i);
    } else {
      cn += max((long long)0, n - x * raise(2, i) + 1);
      break;
    }
  }
  return cn + 1;
}
long long even(long long x) {
  long long cn = 0;
  for (long long i = 1; i <= 64; i++) {
    if (raise(2, i) * x + raise(2, i + 1) - 1 <= n) {
      cn += raise(2, i + 1);
    } else {
      cn += max((long long)0, n - x * raise(2, i) + 1);
      break;
    }
  }
  if (x == n) return cn + 1;
  return cn + 2;
}
void solve() {
  cin >> n >> k;
  long long l = 1, r = n / 2, ans = 0;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (even(2 * mid) >= k) {
      ans = max(ans, 2 * mid);
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  l = 1;
  if (2 * (n / 2) + 1 <= n)
    r = n / 2;
  else
    r = n / 2 - 1;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (odd(2 * mid + 1) >= k) {
      ans = max(ans, 2 * mid + 1);
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  if (odd(1) >= k) ans = max(ans, (long long)1);
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
