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 main() {
  long long n, k, ans1 = 0, ans2 = 0;
  scanf("%lld%*c", &n), scanf("%lld%*c", &k);
  long long low = 2, hi = (n & 1) ? n - 1 : n;
  while (low <= hi) {
    long long mid = (low + hi) / 2, temp = 0, val, point;
    if (mid & 1) {
      mid++;
    }
    temp++;
    if (mid + 1 <= n) temp++;
    val = 2, point = mid;
    while (point < n) {
      point *= 2;
      val *= 2;
      if (point > n) {
        break;
      } else if (point + (val - 1) <= n) {
        temp += val;
      } else {
        temp += (n - point + 1);
      }
    }
    if (temp >= k) {
      ans1 = mid;
      low = (mid + 2);
    } else {
      hi = mid - 2;
    }
  }
  low = 1, hi = n;
  while (low <= hi) {
    long long mid = (low + hi) / 2, temp = 0, val, point;
    if (mid & 1) {
      temp++;
      val = 1, point = mid;
    } else {
      temp++;
      if (mid + 1 <= n) temp++;
      val = 2, point = mid;
    }
    while (point < n) {
      point *= 2;
      val *= 2;
      if (point > n) {
        break;
      } else if (point + (val - 1) <= n) {
        temp += val;
      } else {
        temp += (n - point + 1);
      }
    }
    if (temp >= k) {
      ans2 = mid;
      low = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  printf("%lld\n", max(ans1, ans2));
}
