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;
void printvector(std::vector<long long> v) {
  for (long long i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << "" << endl;
}
void printarray(long long a[], long long n) {
  for (long long i = 0; i < n; ++i) {
    cout << a[i] << " ";
  }
  cout << "" << endl;
}
long long f(long long n, long long x) {
  if (n < x) return 0;
  if (n == x) return 1;
  if (x % 2 == 0) {
    long long left = x, right = x;
    long long sum = 0;
    while (left <= n) {
      sum += (right / 2 - left / 2 + 1);
      left = left * 2;
      right = right * 2 + 2;
      right = min(right, n - n % 2);
    }
    left = x + 1, right = x + 1;
    while (left <= n) {
      sum += ((right - left) / 2 + 1);
      left = left * 2 - 1;
      right = right * 2 + 1;
      right = min(right, n - 1 + n % 2);
    }
    return sum;
  } else {
    return 1 + f(n, 2 * x);
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  if (k == n) {
    cout << 1 << endl;
    return;
  }
  long long ans1 = 1, ans2 = 0;
  long long l = 2, r = n - n % 2;
  while (l < r) {
    long long mid = 2 * ((l / 2 + r / 2) / 2) + 2;
    long long d = f(n, mid);
    if (d >= k) {
      l = mid;
    } else {
      r = mid - 2;
    }
  }
  ans2 = l;
  l = 1;
  r = n - 1 + n % 2;
  while (l < r) {
    long long mid = 2 * ((l / 2 + r / 2) / 2) + 3;
    long long d = f(n, mid);
    if (d >= k) {
      l = mid;
    } else {
      r = mid - 2;
    }
  }
  ans1 = l;
  cout << max(ans2, ans1) << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
