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;
using ll = long long;
ll lim, n, k;
map<ll, ll> cnt;
struct data {
  ll l, r, x, y;
};
vector<data> dp;
ll get(ll x) {
  if (x > lim) return 0;
  if (cnt.count(x)) {
    return cnt[x];
  }
  for (auto it : dp)
    if (it.l <= x && it.r >= x) {
      if (x & 1)
        return it.y;
      else
        return it.x;
    }
}
ll getcnt(ll x) {
  if (x & 1) return 1 + get(2ll * x);
  return 1 + get(2ll * x) + get(x + 1);
}
int main() {
  cin >> n >> k;
  lim = n;
  cnt[n] = 1;
  for (int step = 2;; step++) {
    if (n <= 50) {
      for (ll i = n; i > 0; i--) cnt[i] = getcnt(i);
      break;
    }
    ll x = n / 2;
    while (2 * x <= n) x++;
    for (ll i = n;; i--) {
      cnt[i] = getcnt(i);
      if (i + 2 <= n && cnt[i] == cnt[i + 2]) {
        dp.push_back({x, i + 1, cnt[i], cnt[i + 1]});
        int sz = dp.size();
        if (i & 1) swap(dp[sz - 1].x, dp[sz - 1].y);
        break;
      }
    }
    n = x - 1;
  }
  ll ans = -1;
  for (auto it : cnt) {
    if (it.second >= k) ans = max(ans, it.first);
  }
  cout << ans << "\n";
  return 0;
}
