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;
int i = 1;
long long f[100001], s[100001];
bool ck(long long y) {
  long long p = k;
  if (y % 2) y *= 2, p--;
  i = 1;
  while (s[i] < p) i++;
  long long j = s[i] - p;
  long long l = y + 2, r = n + j + 1;
  i--;
  while (i--) {
    l <<= 1;
    if (l > r) return false;
  }
  return l <= r;
}
long long bs1() {
  long long b = 0, e = n / 2 + 1, cur = (b + e) / 2;
  bool ok = false;
  while (cur > b) {
    ok = ck(2 * cur);
    if (ok) {
      b = cur;
      cur = (b + e) / 2;
    } else {
      e = cur;
      cur = (b + e) / 2;
    }
  }
  return cur;
}
long long bs2() {
  long long b = 1, e = (n + 1) / 2 + 1, cur = (b + e) / 2;
  bool ok = false;
  while (cur > b) {
    ok = ck(2 * cur - 1);
    if (ok) {
      b = cur;
      cur = (b + e) / 2;
    } else {
      e = cur;
      cur = (b + e) / 2;
    }
  }
  return cur;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  f[1] = 2;
  s[1] = 2;
  cin >> n >> k;
  if (k == 1) {
    cout << n << endl;
    return 0;
  }
  while (s[i] < k) {
    i++;
    f[i] = f[i - 1] * 2;
    s[i] = s[i - 1] + f[i];
  }
  cout << max(2 * bs1(), 2 * bs2() - 1) << endl;
  return 0;
}
