#include <bits/stdc++.h>
#define ll long long
using namespace std;
int m,cnt;
vector<int>f;
vector<int>::iterator it;
int main(){
    ll n; cin>>n;
    for (m=40; n<(1ll<<m)-1; --m);
    for (int i=m; i; --i) f.push_back(i);
    for (int i=m; i; --i) f.push_back(i);
    cnt=m,n-=(1ll<<m)-1;
    for (int i=m-1; ~i; --i)
	if (n>>i&1){
	    it=f.end()-i;
	    f.insert(it,++cnt),f.insert(f.begin(),cnt);
	}
    cout<<f.size()<<endl;
    for (int i=0; i<f.size(); ++i)
	cout<<f[i]<<" ";
    return 0;
}

### Prompt

Your task is to create 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 <bits/stdc++.h>
#define ll long long
using namespace std;
int m,cnt;
vector<int>f;
vector<int>::iterator it;
int main(){
    ll n; cin>>n;
    for (m=40; n<(1ll<<m)-1; --m);
    for (int i=m; i; --i) f.push_back(i);
    for (int i=m; i; --i) f.push_back(i);
    cnt=m,n-=(1ll<<m)-1;
    for (int i=m-1; ~i; --i)
	if (n>>i&1){
	    it=f.end()-i;
	    f.insert(it,++cnt),f.insert(f.begin(),cnt);
	}
    cout<<f.size()<<endl;
    for (int i=0; i<f.size(); ++i)
	cout<<f[i]<<" ";
    return 0;
}

```