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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int n, k;
long long int find_ans_odd(long long int num) {
  long long int zz = n / num;
  if (zz == 0) {
    return 0;
  }
  long long int pow = 0, t = 1, pp;
  while (1) {
    if (t > zz) {
      pp = pow - 1;
      break;
    } else {
      pow++;
      t = t * 2;
    }
  }
  long long int ans = 0;
  long long int tmp = 1;
  for (long long int i = 0; i < pp; i++) {
    ans = ans + tmp;
    tmp = tmp * 2;
  }
  long long int zzz = tmp * num;
  long long int count = min(zzz + tmp - 1, n);
  ans = ans + (count - zzz + 1);
  return ans;
}
long long int find_ans_even(long long int num) {
  long long int zz = n / num;
  if (zz == 0) {
    return 0;
  }
  long long int pow = 0, t = 1, pp;
  while (1) {
    if (t > zz) {
      pp = pow - 1;
      break;
    } else {
      pow++;
      t = t * 2;
    }
  }
  long long int ans = 0;
  long long int tmp = 1;
  for (long long int i = 0; i < pp; i++) {
    ans = ans + tmp * 2;
    tmp = tmp * 2;
  }
  long long int zzz = tmp * num;
  long long int count = min(zzz + (tmp * 2) - 1, n);
  ans = ans + (count - zzz + 1);
  return ans;
}
long long int find_odd() {
  long long int low = 1, high = 1000000000000000000 - 1;
  long long int ans = -1;
  long long int mid;
  while (low <= high) {
    mid = (low + high) / 2;
    if (mid % 2 == 0) {
      mid++;
    }
    long long int tmp = find_ans_odd(mid);
    if (low == high) {
      if (tmp >= k) {
        ans = max(ans, mid);
        return ans;
      } else {
        return ans;
      }
    } else if (tmp < k) {
      high = mid - 2;
    } else if (tmp >= k) {
      ans = max(ans, mid);
      low = mid + 2;
    }
  }
  return ans;
}
long long int find_even() {
  long long int low = 2, high = 1000000000000000000;
  long long int ans = -1;
  long long int mid;
  while (low <= high) {
    mid = (low + high) / 2;
    if (mid % 2 == 1) {
      mid++;
    }
    long long int tmp = find_ans_even(mid);
    if (low == high) {
      if (tmp >= k) {
        ans = max(ans, mid);
        return ans;
      } else {
        return ans;
      }
    } else if (tmp < k) {
      high = mid - 2;
    } else if (tmp >= k) {
      ans = max(ans, mid);
      low = mid + 2;
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int T = 1;
  while (T--) {
    cin >> n >> k;
    long long int z = find_odd();
    long long int x = find_even();
    long long int ans = max(z, x);
    cout << ans;
  }
  return 0;
}
