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 long long INF = 1e9;
const long long MAX = 2e5 + 3;
long long MOD = 1e9 + 7;
long long n, k;
long long count_odd(long long x) {
  long long exp = -1;
  long long curr = x;
  while (curr <= n) {
    curr *= 2;
    exp++;
  }
  if (exp == -1) return 0;
  if (curr > n) curr /= 2;
  curr = min(n - curr + 1, curr / x);
  long long ans = 1;
  while (exp) {
    ans *= 2;
    exp--;
  }
  ans--;
  ans += curr;
  return ans;
}
long long count_even(long long x) {
  long long exp = -1;
  long long curr = x;
  while (curr + 2 * curr / x - 1 <= n) {
    curr *= 2;
    exp++;
  }
  if (exp == -1) return curr == n ? 1 : 0;
  if (curr <= n) {
    curr = n - curr + 1;
    exp++;
  } else {
    curr /= 2;
    curr = min(n - curr + 1, 2 * curr / x);
  }
  long long ans = 1;
  while (exp) {
    ans *= 2;
    exp--;
  }
  ans--;
  ans *= 2;
  ans += curr;
  return ans;
}
int32_t main() {
  cin >> n >> k;
  long long oddans = 0, evenans = 0;
  long long lo = 1, hi = n / 2 + 2, mid;
  while (lo <= hi) {
    mid = (lo + hi) / 2;
    long long codd = count_odd(2 * (mid - 1) + 1);
    if (codd < k)
      hi = mid - 1;
    else if (codd >= k) {
      lo = mid + 1;
      oddans = 2 * (mid - 1) + 1;
    }
  }
  lo = 1, hi = n / 2 + 2;
  while (lo <= hi) {
    mid = (lo + hi) / 2;
    long long ceven = count_even(2 * mid);
    if (ceven < k)
      hi = mid - 1;
    else if (ceven >= k) {
      lo = mid + 1;
      evenans = 2 * mid;
    }
  }
  cout << max(oddans, evenans);
}
