#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
long long n, k;
int check(long long x) {
  long long ans = 0, a, b;
  if (x % 2 == 1 && x <= n) {
    ans += 1;
    a = x, b = x;
    while (1) {
      a = a * 2;
      b = b * 2 + 1;
      if (b <= n)
        ans += b - a + 1;
      else {
        if (a > n)
          break;
        else {
          ans += n - a + 1;
          break;
        }
      }
    }
    if (ans >= k) return 1;
    return 0;
  }
  if (x == n - 1) return 2 >= k;
  if (x % 2 == 0 && x <= n) {
    ans += 2;
    a = x, b = x + 1;
    while (1) {
      a = a * 2;
      b = b * 2 + 1;
      if (b <= n)
        ans += b - a + 1;
      else {
        if (a > n)
          break;
        else {
          ans += n - a + 1;
          break;
        }
      }
    }
  }
  if (ans >= k) return 1;
  return 0;
}
int main() {
  scanf("%lld%lld", &n, &k);
  long long l = 1, r = n, mid, ans = 0;
  while (l <= r) {
    mid = (l + r) / 2;
    if (check(2 * mid)) {
      l = mid + 1;
    } else
      r = mid - 1;
  }
  ans = max(ans, (l - 1) * 2);
  l = 1, r = n, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (check(2 * mid - 1)) {
      l = mid + 1;
    } else
      r = mid - 1;
  }
  ans = max(ans, (l - 1) * 2 - 1);
  printf("%lld\n", ans);
}

### 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;
const int inf = 0x3f3f3f3f;
long long n, k;
int check(long long x) {
  long long ans = 0, a, b;
  if (x % 2 == 1 && x <= n) {
    ans += 1;
    a = x, b = x;
    while (1) {
      a = a * 2;
      b = b * 2 + 1;
      if (b <= n)
        ans += b - a + 1;
      else {
        if (a > n)
          break;
        else {
          ans += n - a + 1;
          break;
        }
      }
    }
    if (ans >= k) return 1;
    return 0;
  }
  if (x == n - 1) return 2 >= k;
  if (x % 2 == 0 && x <= n) {
    ans += 2;
    a = x, b = x + 1;
    while (1) {
      a = a * 2;
      b = b * 2 + 1;
      if (b <= n)
        ans += b - a + 1;
      else {
        if (a > n)
          break;
        else {
          ans += n - a + 1;
          break;
        }
      }
    }
  }
  if (ans >= k) return 1;
  return 0;
}
int main() {
  scanf("%lld%lld", &n, &k);
  long long l = 1, r = n, mid, ans = 0;
  while (l <= r) {
    mid = (l + r) / 2;
    if (check(2 * mid)) {
      l = mid + 1;
    } else
      r = mid - 1;
  }
  ans = max(ans, (l - 1) * 2);
  l = 1, r = n, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (check(2 * mid - 1)) {
      l = mid + 1;
    } else
      r = mid - 1;
  }
  ans = max(ans, (l - 1) * 2 - 1);
  printf("%lld\n", ans);
}

```