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;
const int num = 200001;
vector<long long> vect;
long long n, k;
bool check(long long mid) {
  long long cnt = 0;
  for (long long nu = n; nu > 0; nu = nu / 2) cnt++;
  long long ans = 0;
  if (mid & 1) {
    for (int i = 1; i < cnt + 1; i++) {
      long long val = vect[i] * mid + vect[i] - 1;
      if (val <= n)
        ans += vect[i];
      else {
        if (n >= (vect[i] * mid)) ans += (n - (vect[i] * mid) + 1);
        break;
      }
    }
    ans++;
  } else {
    for (int i = 1; i < cnt + 1; i++) {
      long long val = 2 * vect[i];
      long long val1 = vect[i] * mid + val - 1;
      if (val1 <= n)
        ans += val;
      else {
        if (n >= (vect[i] * mid)) ans += (n - vect[i] * mid + 1);
        break;
      }
    }
    ans += 2;
  }
  if (ans >= k)
    return true;
  else
    return false;
}
long long bs(long long low, long long high, long long ptr) {
  long long l = (low - ptr) / 2;
  long long h = (high - ptr) / 2;
  while ((l + 1) < h) {
    long long m = (l + h) / 2;
    if (check(2 * m + ptr))
      l = m;
    else
      h = m;
  }
  if (check(2 * h + ptr))
    return (2 * h + ptr);
  else
    return (2 * l + ptr);
}
int main() {
  vect.push_back(1);
  for (int i = 1; i < 60; i++) {
    vect.push_back(vect[i - 1] * 2ll);
  }
  cin >> n >> k;
  if (n == k) {
    cout << 1 << "\n";
    return 0;
  }
  long long low = 1;
  long long high = n;
  long long mid;
  if (n & 1) {
    cout << max(bs(2, n - 1, 0), bs(1, n, 1)) << "\n";
  } else
    cout << max(bs(2, n, 0), bs(1, n - 1, 1)) << "\n";
  return 0;
}
