#include <bits/stdc++.h>
using namespace std;
unsigned long long n, k;
int getlen_b(unsigned long long k) {
  int len = 0;
  for (; k; k >>= 1) len++;
  return len;
}
void print_binary(unsigned long long n) {
  int sta[100], top = 0;
  for (; n; n >>= 1) sta[++top] = n & 1;
  for (int i = top; i; i--) cout << sta[i];
  cout << endl;
}
unsigned long long calc(unsigned long long k, unsigned long long n) {
  int lenk = getlen_b(k) - 1, lenn = getlen_b(n);
  int i;
  unsigned long long res = k & 1;
  if (k & 1)
    lenk++;
  else
    k >>= 1;
  unsigned long long nn = n >> (lenn - lenk);
  for (i = 1; lenk + i < lenn; ++i) res += (1ll << i);
  if (k < nn)
    res += 1ll << (lenn - lenk);
  else if (k == nn)
    res += n - (k << (lenn - lenk)) + 1;
  return res;
}
void work() {
  cin >> n >> k;
  long long l = 1, r = n / 2, mid, res1 = -1, res2 = -1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (calc(mid * 2, n) >= k)
      l = mid + 1, res2 = mid * 2;
    else
      r = mid - 1;
  }
  l = 0, r = (n - 1) / 2;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (calc(mid * 2 + 1, n) >= k)
      l = mid + 1, res1 = mid * 2 + 1;
    else
      r = mid - 1;
  }
  cout << max(res1, res2) << endl;
  return;
}
int main() {
  work();
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
unsigned long long n, k;
int getlen_b(unsigned long long k) {
  int len = 0;
  for (; k; k >>= 1) len++;
  return len;
}
void print_binary(unsigned long long n) {
  int sta[100], top = 0;
  for (; n; n >>= 1) sta[++top] = n & 1;
  for (int i = top; i; i--) cout << sta[i];
  cout << endl;
}
unsigned long long calc(unsigned long long k, unsigned long long n) {
  int lenk = getlen_b(k) - 1, lenn = getlen_b(n);
  int i;
  unsigned long long res = k & 1;
  if (k & 1)
    lenk++;
  else
    k >>= 1;
  unsigned long long nn = n >> (lenn - lenk);
  for (i = 1; lenk + i < lenn; ++i) res += (1ll << i);
  if (k < nn)
    res += 1ll << (lenn - lenk);
  else if (k == nn)
    res += n - (k << (lenn - lenk)) + 1;
  return res;
}
void work() {
  cin >> n >> k;
  long long l = 1, r = n / 2, mid, res1 = -1, res2 = -1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (calc(mid * 2, n) >= k)
      l = mid + 1, res2 = mid * 2;
    else
      r = mid - 1;
  }
  l = 0, r = (n - 1) / 2;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (calc(mid * 2 + 1, n) >= k)
      l = mid + 1, res1 = mid * 2 + 1;
    else
      r = mid - 1;
  }
  cout << max(res1, res2) << endl;
  return;
}
int main() {
  work();
  return 0;
}

```