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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  auto count = [&](long long x) {
    if (x % 2 == 1) {
      long long left = x, right = x;
      long long ans = 0;
      while (left <= n) {
        ans += min(right, n) - left + 1;
        left = left * 2;
        right = right * 2 + 1;
      }
      return ans;
    } else {
      long long left = x, right = x + 1;
      long long ans = 0;
      while (left <= n) {
        ans += min(right, n) - left + 1;
        left = left * 2;
        right = right * 2 + 1;
      }
      return ans;
    }
  };
  cerr << count(2) << endl;
  cerr << count(3) << endl;
  long long ans = 1;
  {
    long long low = 1, high = n % 2 == 0 ? n - 1 : n;
    while (low <= high) {
      long long mid = (low + high) / 2;
      if (mid % 2 == 0) {
        if (low < mid) {
          mid--;
        } else {
          mid++;
        }
      }
      long long c = count(mid);
      if (c >= k) {
        ans = max(ans, mid);
        low = mid + 2;
      } else {
        high = mid - 2;
      }
    }
  }
  {
    long long low = 2, high = n % 2 == 0 ? n : n - 1;
    while (low <= high) {
      long long mid = (low + high) / 2;
      if (mid % 2 == 1) {
        if (low < mid) {
          mid--;
        } else {
          mid++;
        }
      }
      long long c = count(mid);
      if (c >= k) {
        ans = max(ans, mid);
        low = mid + 2;
      } else {
        high = mid - 2;
      }
    }
  }
  cout << ans << '\n';
}
