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 n, k;
long long get(long long x) {
  assert(x % 2 == 0);
  long long cnt = 0;
  for (long long i = 0;; i++) {
    if (x > n / (1LL << i)) break;
    long long kk = x * (1LL << i);
    long long Pow = 1LL << (i + 1);
    long long R = min(Pow - 1, n - kk);
    cnt += R + 1;
  }
  return cnt;
}
long long bsOdd() {
  long long res = -1;
  for (long long low = 0, high = (n - 1) / 2; low <= high;) {
    long long mid = (low + high) >> 1;
    long long val = 2 * mid + 1;
    assert(val <= n);
    if (get(2 * val) + 1 >= k)
      res = val, low = mid + 1;
    else
      high = mid - 1;
  }
  return res;
}
long long bsEven() {
  long long res = -1;
  for (long long low = 1, high = n / 2; low <= high;) {
    long long mid = (low + high) >> 1;
    long long val = 2 * mid;
    if (get(val) >= k)
      res = val, low = mid + 1;
    else
      high = mid - 1;
  }
  return res;
}
int32_t main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("main.in", "r"))
    freopen("main.in", "r", stdin), freopen("main.out", "w", stdout);
  cin >> n >> k;
  long long res1 = bsOdd();
  long long res2 = bsEven();
  cout << max(res1, res2) << '\n';
  return 0;
}
