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;
int xm[4] = {-1, 1, 0, 0};
int ym[4] = {0, 0, -1, 1};
const int MOD = 1e9 + 7;
const int MAXN = 5e5 + 5;
const long long POW = 9973;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  long long l = 1;
  long long r = n;
  long long ans = 1;
  while (l <= r) {
    long long m = l + (r - l) / 2;
    long long res = 0;
    for (long long st = m; st <= m + 1; st++) {
      if (st <= 0 || st > n) continue;
      long long cnt = 0;
      long long cur = st;
      long long top = st;
      while (cur <= n) {
        if (top % 2 == 0) top++;
        cnt += min(n, top) - cur + 1;
        cur *= 2;
        top *= 2;
      }
      if (cnt >= k) res = st;
    }
    if (res != 0) {
      ans = max(ans, res);
      l = res + 1;
    } else
      r = m - 1;
  }
  cout << ans;
}
