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;
unsigned long long a[400000], b[400000];
unsigned long long bit(unsigned long long n) {
  unsigned long long cnt = 0;
  while (n) {
    cnt++;
    n /= 2;
  }
  return cnt;
}
unsigned long long f(unsigned long long n, unsigned long long N) {
  if (n == 0) return 0;
  unsigned long long l1, l2, n1, j = 0, ans = 0, m = n, k, p = 1;
  l1 = bit(N);
  if (m % 2 == 0) n /= 2;
  while (bit(n) < l1) {
    n *= 2;
    ans += (p << j);
    j++;
  }
  n1 = N;
  n1 = n1 >> j;
  n1 = n1 << j;
  if (n1 > n)
    k = ans + (p << j);
  else if (n1 < n)
    k = ans;
  else {
    k = ans + N - n1 + 1;
  }
  if (m % 2)
    return k;
  else
    return k - 1;
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
  };
  unsigned long long i, j, k, l = 0, n, m, h, mod = 1e9 + 7, cnt = 0, mx = 1e10,
                              t, ans = 0;
  cin >> n >> k;
  j = n / 2;
  l = 0;
  h = j;
  while (h - l > 1) {
    m = (l + h) / 2;
    if (f(2 * m, n) < k)
      h = m - 1;
    else
      l = m;
  }
  if (f(2 * h, n) >= k) l = h;
  t = 2 * l;
  l = 1;
  h = j;
  if (n % 2) h++;
  while (h - l > 1) {
    m = (l + h) / 2;
    if (f(2 * m - 1, n) < k)
      h = m - 1;
    else
      l = m;
  }
  if (f(2 * h - 1, n) + 1 >= k) l = h;
  cout << max(2 * l - 1, t) << endl;
  return 0;
}
