#include <algorithm>
#include <cstdio>
#include <vector>
using namespace std;
typedef long long int64;
int64 n;
vector<int> meth[2];
int m;
int main()
{
    scanf("%lld", &n), ++n;
    for(int i = 1; n > 1; ++i) meth[n&1].push_back(i), n = n&1?(n-1):(n>>1);
    m = meth[0].size()+meth[1].size(); printf("%d\n", m<<1);    
    for(int i = 0; i < meth[1].size(); ++i) printf("%d ", meth[1][i]);
    for(int i = meth[0].size()-1; ~i; --i) printf("%d ", meth[0][i]);
    for(int i = 1; i <= m; ++i) printf("%d ", m-i+1);
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 <algorithm>
#include <cstdio>
#include <vector>
using namespace std;
typedef long long int64;
int64 n;
vector<int> meth[2];
int m;
int main()
{
    scanf("%lld", &n), ++n;
    for(int i = 1; n > 1; ++i) meth[n&1].push_back(i), n = n&1?(n-1):(n>>1);
    m = meth[0].size()+meth[1].size(); printf("%d\n", m<<1);    
    for(int i = 0; i < meth[1].size(); ++i) printf("%d ", meth[1][i]);
    for(int i = meth[0].size()-1; ~i; --i) printf("%d ", meth[0][i]);
    for(int i = 1; i <= m; ++i) printf("%d ", m-i+1);
    return 0;
}
```