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 int const LINF = numeric_limits<long long int>::max();
int const INF = numeric_limits<int>::max();
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
long long int pow_mod(long long int a, long long int e, long long int m) {
  long long int res = 1;
  while (e) {
    if (e & 1) res = res * a % m;
    a = a * a % m;
    e >>= 1;
  }
  return res;
}
long long int inv_mod(long long int b, long long int m) {
  b %= m;
  long long int x = 0, y = 1, r, q, a = m;
  while (b) {
    q = a / b;
    r = a % b;
    a = b;
    b = r;
    r = x;
    x = y;
    y = r - q * y;
  }
  x += m;
  return x % m;
}
long long int bs(long long int s, long long int p, long long int n) {
  long long int l = 0, h = p - 1;
  while (l < h) {
    long long int m = (l + h) >> 1;
    if (s + m <= n)
      l = m + 1;
    else
      h = m;
  }
  if (s + l > n) l--;
  return l + 1;
}
long long int paths(long long int m, long long int n) {
  long long int y = m, cnt = 0, p = 2;
  while (y <= n) {
    cnt += bs(y, p, n);
    y <<= 1;
    p <<= 1;
  }
  return cnt;
}
long long int solve_even(long long int n, long long int k) {
  long long int l = 1, h = n / 2;
  while (l < h) {
    long long int m = (l + h) >> 1, cnt = paths(2 * m, n);
    if (cnt < k)
      h = m;
    else
      l = m + 1;
  }
  long long int cnt = paths(2 * l, n);
  if (cnt < k) l--;
  return l * 2;
}
long long int solve_odd(long long int n, long long int k) {
  long long int l = 1, h = n / 2 + (n & 1);
  while (l < h) {
    long long int m = (l + h) >> 1, cnt = paths(4 * m - 2, n) + 1;
    if (cnt < k)
      h = m;
    else
      l = m + 1;
  }
  long long int cnt = paths(4 * l - 2, n) + 1;
  if (cnt < k) l--;
  return l * 2 - 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int n, k;
  cin >> n >> k;
  long long int even = solve_even(n, k);
  long long int odd = solve_odd(n, k);
  cout << max(even, odd) << endl;
}
