#include <iostream>

using namespace std;

typedef long long LL;
int a[103];
int b[103];

int main()
{
    LL n;
    cin>>n;
    n++;

    int m=100;
    int i=0,j=0;
    while(n>1)
    {
        if(n&1) {a[i++]=m--;n--;}
        else {b[j++]=m--;n>>=1;}
    }

    cout<<2*(i+j)<<endl;
    for(int k=0;k<i;k++) cout<<a[k]<<' ';
    for(int k=j-1;k>=0;k--) cout<<b[k]<<' ';
    for(int k=m+1;k<=100;k++) cout<<k<<' ';

    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 <iostream>

using namespace std;

typedef long long LL;
int a[103];
int b[103];

int main()
{
    LL n;
    cin>>n;
    n++;

    int m=100;
    int i=0,j=0;
    while(n>1)
    {
        if(n&1) {a[i++]=m--;n--;}
        else {b[j++]=m--;n>>=1;}
    }

    cout<<2*(i+j)<<endl;
    for(int k=0;k<i;k++) cout<<a[k]<<' ';
    for(int k=j-1;k>=0;k--) cout<<b[k]<<' ';
    for(int k=m+1;k<=100;k++) cout<<k<<' ';

    return 0;
}
```