#include<cstdio>
#include<deque>
using namespace std;
typedef long long LL;
int t=1;
deque<int> q;
void dfs(LL n)
{
    if(n==1) return;
    if((n-1)%2==0)
    {
        dfs((n-1)/2);
        q.push_back(++t);
    }
    else
    {
        dfs(n-1);
        q.push_front(++t);
    }
}
int main()
{
    LL n;
    int i,m;
    scanf("%lld",&n);
    q.push_back(1);
    dfs(n);
    m=q.size();
    printf("%d\n",m*2);
    while(!q.empty())
    {
        printf("%d ",q.front());
        q.pop_front();
    }
    for(i=1;i<=m;i++)
        printf("%d ",i);
    printf("\n");
    return 0;
}
### Prompt

Please formulate a CPP solution to the following problem:

We will call a string x good if it satisfies the following condition:

* Condition: x can be represented as a concatenation of two copies of another string y of length at least 1.



For example, `aa` and `bubobubo` are good; an empty string, `a`, `abcabcabc` and `abba` are not good.

Eagle and Owl created a puzzle on good strings. Find one string s that satisfies the following conditions. It can be proved that such a string always exists under the constraints in this problem.

* 1 ≤ |s| ≤ 200
* Each character of s is one of the 100 characters represented by the integers 1 through 100.
* Among the 2^{|s|} subsequences of s, exactly N are good strings.

Constraints

* 1 ≤ N ≤ 10^{12}

Input

Input is given from Standard Input in the following format:


N


Output

In the first line, print |s|, the length of s. In the second line, print the elements in s in order, with spaces in between. Any string that satisfies the above conditions will be accepted.

Examples

Input

7


Output

4
1 1 1 1


Input

299


Output

23
32 11 11 73 45 8 11 83 83 8 45 32 32 10 100 73 32 83 45 73 32 11 10

### Response

```cpp
#include<cstdio>
#include<deque>
using namespace std;
typedef long long LL;
int t=1;
deque<int> q;
void dfs(LL n)
{
    if(n==1) return;
    if((n-1)%2==0)
    {
        dfs((n-1)/2);
        q.push_back(++t);
    }
    else
    {
        dfs(n-1);
        q.push_front(++t);
    }
}
int main()
{
    LL n;
    int i,m;
    scanf("%lld",&n);
    q.push_back(1);
    dfs(n);
    m=q.size();
    printf("%d\n",m*2);
    while(!q.empty())
    {
        printf("%d ",q.front());
        q.pop_front();
    }
    for(i=1;i<=m;i++)
        printf("%d ",i);
    printf("\n");
    return 0;
}
```