#include <bits/stdc++.h>
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; c > '9' || c < '0'; c = getchar()) {
    if (f == '-') f = -1;
  }
  for (; c >= '0' && c <= '9'; c = getchar()) {
    x = x * 10 + c - '0';
  }
  return x * f;
}
inline bool check(long long x, long long n, long long k) {
  long long res = 0, l = x, r = x;
  if (!(x & 1)) r++;
  for (;;) {
    res += std::min(n, r) - l + 1;
    l <<= 1;
    r = (r << 1) + 1;
    if (l > n) break;
  }
  return res >= k;
}
inline void work() {
  long long n = read(), k = read(), ans = 0;
  for (long long l = 1, r = n + 1 >> 1; l <= r;) {
    long long mid = l + r >> 1;
    if (check((mid << 1) - 1, n, k)) {
      ans = (mid << 1) - 1;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  for (long long l = 1, r = n >> 1; l <= r;) {
    long long mid = l + r >> 1;
    if (check(mid << 1, n, k)) {
      ans = std::max(ans, mid << 1);
      l = mid + 1;
    } else
      r = mid - 1;
  }
  printf("%I64d\n", ans);
}
int32_t main() {
  work();
  return 0;
}

### Prompt

Your challenge is to write 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>
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; c > '9' || c < '0'; c = getchar()) {
    if (f == '-') f = -1;
  }
  for (; c >= '0' && c <= '9'; c = getchar()) {
    x = x * 10 + c - '0';
  }
  return x * f;
}
inline bool check(long long x, long long n, long long k) {
  long long res = 0, l = x, r = x;
  if (!(x & 1)) r++;
  for (;;) {
    res += std::min(n, r) - l + 1;
    l <<= 1;
    r = (r << 1) + 1;
    if (l > n) break;
  }
  return res >= k;
}
inline void work() {
  long long n = read(), k = read(), ans = 0;
  for (long long l = 1, r = n + 1 >> 1; l <= r;) {
    long long mid = l + r >> 1;
    if (check((mid << 1) - 1, n, k)) {
      ans = (mid << 1) - 1;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  for (long long l = 1, r = n >> 1; l <= r;) {
    long long mid = l + r >> 1;
    if (check(mid << 1, n, k)) {
      ans = std::max(ans, mid << 1);
      l = mid + 1;
    } else
      r = mid - 1;
  }
  printf("%I64d\n", ans);
}
int32_t main() {
  work();
  return 0;
}

```