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;
inline int _lg(long long x) { return 63 - __builtin_clzll(x); }
long long n, k, ans;
inline long long calc(long long x) {
  if (x <= 0 || x > n) return 0;
  int bn = _lg(n);
  int bx = _lg(x);
  long long o = x - (n >> (bn - bx)), num = 0;
  if (o > 0)
    num = (1ll << (bn - bx)) - 1;
  else if (o < 0)
    num = (2ll << (bn - bx)) - 1;
  else
    num = (n & ((1ll << (bn - bx)) - 1)) + (1ll << (bn - bx));
  if (x % 2 == 0) num += calc(x + 1);
  return num;
}
inline void upd(long long x) {
  long long cnt = calc(x);
  if (cnt >= k && x >= ans) ans = x;
}
int main() {
  scanf("%lld%lld", &n, &k);
  for (int i = 1; i <= (100); ++i) upd(i);
  for (int i = 0; i <= (61); ++i)
    upd((1ll << i) - 1), upd((1ll << i) - 2), upd((1ll << i) - 3);
  long long m = n;
  while (m)
    upd(m + 3), upd(m + 2), upd(m + 1), upd(m), upd(m - 1), upd(m - 2),
        upd(m - 3), m >>= 1;
  printf("%lld\n", ans);
  return 0;
}
