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 binarySearch(long long low, long long high, long long n,
                       long long flag, long long k) {
  long long ans = 0;
  while (low <= high) {
    if (low == high && low % 2 == flag) break;
    long long mid = (low + high) / 2;
    if (flag && mid % 2)
      mid = mid + 1;
    else if (flag == 0 && mid % 2 == 0)
      mid = mid + 1;
    long long sum = 0;
    if (mid % 2) {
      long long term = log2(n / mid);
      if (term > 0) sum = (1ll << (term)) - 1;
      long long lastTerm = 1ll << (term);
      long long prod = mid * (1ll << term);
      if (n - prod + 1 <= lastTerm)
        sum += n - prod + 1;
      else
        sum += lastTerm;
    } else {
      long long term = log2(n / mid);
      if (term > 0) sum = 2 * ((1ll << (term)) - 1);
      long long lastTerm = 2 * (1ll << (term));
      long long prod = mid * (1ll << term);
      if (n - prod + 1 <= lastTerm)
        sum += n - prod + 1;
      else
        sum += lastTerm;
    }
    if (sum >= k) ans = max(ans, mid);
    if (sum < k)
      high = mid - 1;
    else
      low = mid + 1;
  }
  return ans;
}
int main() {
  long long int n, k;
  cin >> n >> k;
  long long low = 1, high = n;
  long long even = binarySearch(1, n, n, 1, k);
  long long odd = binarySearch(1, n, n, 0, k);
  cout << max(even, odd) << "\n";
}
