#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef long long ll;
ll n;int l,r;
vector<int> Q(ll n){
    vector<int> ans;
    if (n<=2){
        for (int i=1;i<=n;i++) ans.push_back(--l);
        return ans;
    }
    ans=Q((n-1)>>1);
    ans.push_back(++r);
    if (!(n&1)) ans.push_back(--l);
    return ans;
}
vector<int> a;
int main()
{
    scanf("%lld",&n);
    l=51;r=50;
    a=Q(n);
    for (int i=l;i<=r;i++) a.push_back(i);
    printf("%d\n",(int)a.size());
    for (int i=0;i<a.size();i++) printf("%d ",a[i]);
    puts("");
    return 0;
}
### Prompt

Create a solution in CPP for 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<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
typedef long long ll;
ll n;int l,r;
vector<int> Q(ll n){
    vector<int> ans;
    if (n<=2){
        for (int i=1;i<=n;i++) ans.push_back(--l);
        return ans;
    }
    ans=Q((n-1)>>1);
    ans.push_back(++r);
    if (!(n&1)) ans.push_back(--l);
    return ans;
}
vector<int> a;
int main()
{
    scanf("%lld",&n);
    l=51;r=50;
    a=Q(n);
    for (int i=l;i<=r;i++) a.push_back(i);
    printf("%d\n",(int)a.size());
    for (int i=0;i<a.size();i++) printf("%d ",a[i]);
    puts("");
    return 0;
}
```