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 add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) {
    a -= 1000000007;
  }
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) {
    a += 1000000007;
  }
  return a;
}
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
void add_self(long long& a, long long b) { a = add(a, b); }
void sub_self(long long& a, long long b) { a = sub(a, b); }
void mul_self(long long& a, long long b) { a = mul(a, b); }
const long long MAXN = 200010;
long long get(long long val, long long tot) {
  long long ans = 0;
  long long L = val, R = val + 1 - (val % 2);
  while (R <= tot) {
    ans += R - L + 1;
    L = 2 * L;
    R = 2 * R + 1;
  }
  ans += max(0LL, tot - L + 1);
  return ans;
}
int main() {
  long long x, k;
  cin >> x >> k;
  long long lo = 0, hi = 1e18, ans = 0;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    if (get(2 * mid + 1, x) >= k) {
      ans = 2 * mid + 1;
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  lo = 0;
  hi = 1e18;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    if (get(2 * mid, x) >= k) {
      ans = max(ans, 2 * mid);
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  cout << ans << "\n";
}
