#include <bits/stdc++.h>
using namespace std;
long long n, k, len, cnt, ans;
long long calc(long long x) {
  long long tx = x;
  len = 0;
  while (tx) {
    tx >>= 1;
    len++;
  }
  long long tn = n >> (cnt - len), ret = 1;
  if (tn == x) {
    ret += (((1ll << (cnt - len)) - 1) & n) + 1;
  } else if (tn > x) {
    ret += (1ll << (cnt - len));
  }
  for (long long i = len + 1; i < cnt; i++) ret += (1ll << (i - len));
  return ret + ((x % 2 == 0) ? calc(x + 1) : 0);
}
signed main() {
  scanf("%lld%lld", &n, &k);
  long long tn = n;
  while (tn) {
    tn >>= 1;
    cnt++;
  }
  long long l = 0, r = (n >> 1) << 1;
  while (l + 2 < r) {
    long long realmid;
    if (((l + r) >> 1) & 1)
      realmid = ((l + r) >> 1) - 1;
    else
      realmid = ((l + r) >> 1);
    if (calc(realmid) >= k)
      l = realmid;
    else
      r = realmid - 2;
  }
  if (calc(r) >= k)
    ans = r;
  else
    ans = l;
  l = 1, r = (n % 2 == 0 ? n - 1 : n);
  while (l + 2 < r) {
    long long realmid;
    if (((l + r) >> 1) & 1)
      realmid = ((l + r) >> 1);
    else
      realmid = ((l + r) >> 1) - 1;
    if (calc(realmid) >= k)
      l = realmid;
    else
      r = realmid - 2;
  }
  if (calc(r) >= k)
    ans = max(r, ans);
  else
    ans = max(l, ans);
  printf("%lld", ans);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
long long n, k, len, cnt, ans;
long long calc(long long x) {
  long long tx = x;
  len = 0;
  while (tx) {
    tx >>= 1;
    len++;
  }
  long long tn = n >> (cnt - len), ret = 1;
  if (tn == x) {
    ret += (((1ll << (cnt - len)) - 1) & n) + 1;
  } else if (tn > x) {
    ret += (1ll << (cnt - len));
  }
  for (long long i = len + 1; i < cnt; i++) ret += (1ll << (i - len));
  return ret + ((x % 2 == 0) ? calc(x + 1) : 0);
}
signed main() {
  scanf("%lld%lld", &n, &k);
  long long tn = n;
  while (tn) {
    tn >>= 1;
    cnt++;
  }
  long long l = 0, r = (n >> 1) << 1;
  while (l + 2 < r) {
    long long realmid;
    if (((l + r) >> 1) & 1)
      realmid = ((l + r) >> 1) - 1;
    else
      realmid = ((l + r) >> 1);
    if (calc(realmid) >= k)
      l = realmid;
    else
      r = realmid - 2;
  }
  if (calc(r) >= k)
    ans = r;
  else
    ans = l;
  l = 1, r = (n % 2 == 0 ? n - 1 : n);
  while (l + 2 < r) {
    long long realmid;
    if (((l + r) >> 1) & 1)
      realmid = ((l + r) >> 1);
    else
      realmid = ((l + r) >> 1) - 1;
    if (calc(realmid) >= k)
      l = realmid;
    else
      r = realmid - 2;
  }
  if (calc(r) >= k)
    ans = max(r, ans);
  else
    ans = max(l, ans);
  printf("%lld", ans);
  return 0;
}

```