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 T>
inline void read(T &x) {
  char c = getchar();
  int w = 1;
  x = 0;
  while (!isdigit(c)) (c == '-') && (w = -w), c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ '0'), c = getchar();
  x *= w;
}
const int dx[8] = {-1, 1, 0, 0, -1, 1, -1, 1},
          dy[8] = {0, 0, -1, 1, -1, -1, 1, 1};
signed long long n, k;
inline signed long long check(signed long long x) {
  signed long long res = 1, p = x;
  x <<= 1;
  if (x > n) {
    if (p % 2 == 0) res += check(p + 1);
    return res;
  }
  int w = 0;
  while (1) {
    x *= 2;
    ++w;
    if (x <= n)
      res += 1ll << w;
    else {
      x >>= 1;
      signed long long y = 1;
      for (register int i = w - 1; ~i; --i)
        if (x + (1ll << i) <= n) x += 1ll << i, y += 1LL << i;
      res += y;
      if (p % 2 == 0) res += check(p + 1);
      return res;
    }
  }
}
int main() {
  read(n);
  read(k);
  signed long long l = 0, r = n / 2 - (1 - n % 2);
  while (l <= r) {
    signed long long mid = ((r - l + 1) >> 1) + l, kk = (mid << 1) + 1;
    if (check(kk) >= k)
      l = mid + 1;
    else
      r = mid - 1;
  }
  signed long long ans = (r << 1) + 1;
  l = 1, r = n / 2;
  while (l <= r) {
    signed long long mid = ((r - l + 1) >> 1) + l, kk = mid << 1;
    if (check(kk) >= k)
      l = mid + 1;
    else
      r = mid - 1;
  }
  cout << max(ans, (r << 1));
  getchar();
  return 0;
}
