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 test = 0;
const int MAXN = 200009;
const long long MOD = 998244353;
class {
 public:
  int len;
  long long n, k;
  void solve() {
    cin >> n >> k;
    if (n == k) {
      cout << 1;
      return;
    }
    for (int i = 60; i >= 0; i--) {
      if (n & (1ll << i)) {
        len = i + 1;
        break;
      }
    }
    long long cur = 1, len = 0;
    while (1) {
      if (call(cur << 1) < k) {
        break;
      }
      cur <<= 1;
      len++;
    }
    for (int i = len - 1; i >= 0; --i) {
      if (call(cur | (1ll << i)) >= k) {
        cur |= 1ll << i;
      }
    }
    cout << cur << '\n';
  }
  long long call(long long x) {
    if (x & 1ll) return cal(x);
    return cal(x + 1) + cal(x);
  }
  long long cal(long long x) {
    int xlen;
    for (int i = 60; i >= 0; i--) {
      if (x & (1ll << i)) {
        xlen = i + 1;
        break;
      }
    }
    long long res = 0, p = 1;
    while (xlen < len) {
      res += p;
      p <<= 1;
      xlen++;
      x <<= 1;
    }
    if (x <= n) {
      res += min(n - x + 1, p);
    }
    return res;
  }
} NSPACE;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}
