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 int const mod = 1000000007;
long long int gcdfun(long long int a, long long int b) {
  if (b == 0) return a;
  a %= b;
  return gcdfun(b, a);
}
long long int power(long long int x, long long int y, long long int m) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2, m) % m;
  if (y % 2 == 0)
    return ((temp) * (temp)) % m;
  else
    return (((x) % m) * ((temp * temp) % m)) % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  long long int lo = 1;
  long long int hi = n;
  while (lo + 4 < hi) {
    long long int mid = lo + (hi - lo) / 2;
    long long int num = 0;
    if (mid % 2 && mid > 1) mid--;
    num = 2;
    long long int temp = mid * 2;
    long long int cnt;
    if (mid % 2)
      cnt = 2;
    else
      cnt = 4;
    while (temp <= n) {
      num += (min(cnt, n - temp + 1));
      cnt *= 2;
      temp *= 2;
    }
    if (num >= k) {
      lo = mid;
    } else {
      hi = mid - 1;
    }
  }
  for (long long int i = hi; i >= lo; i--) {
    if (i > n) continue;
    long long int num = 0;
    if (i % 2)
      num = 1;
    else
      num = 2;
    long long int temp = i * 2;
    long long int cnt;
    if (i % 2)
      cnt = 2;
    else
      cnt = 4;
    while (temp <= n) {
      num += (min(cnt, n - temp + 1));
      cnt *= 2;
      temp *= 2;
    }
    if (num >= k) {
      cout << i << "\n";
      return 0;
    }
  }
}
