#include<iostream>
#include<algorithm>
using namespace std;
long long n;
int p_1[106],p_0[106];

int main(){
    cin>>n;
    n++;
    int k1=0,k2=0;
    int m=100;
    while(n>1){
        if((n&1)==1){
            p_1[k1]=m--;k1++;n--;
        }
        else{
            p_0[k2]=m--;k2++;n=n>>1;
        }
    }
    cout<<(k1+k2)*2<<endl;
    for(int i=0;i<k1;i++)
        cout<<p_1[i]<<" ";
    for(int i=k2-1;i>=0;i--)
        cout<<p_0[i]<<" ";
    int p=101-k1-k2;
    for(;p<100;p++)
        cout<<p<<" ";
    cout<<"100"<<endl;
    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
#include<algorithm>
using namespace std;
long long n;
int p_1[106],p_0[106];

int main(){
    cin>>n;
    n++;
    int k1=0,k2=0;
    int m=100;
    while(n>1){
        if((n&1)==1){
            p_1[k1]=m--;k1++;n--;
        }
        else{
            p_0[k2]=m--;k2++;n=n>>1;
        }
    }
    cout<<(k1+k2)*2<<endl;
    for(int i=0;i<k1;i++)
        cout<<p_1[i]<<" ";
    for(int i=k2-1;i>=0;i--)
        cout<<p_0[i]<<" ";
    int p=101-k1-k2;
    for(;p<100;p++)
        cout<<p<<" ";
    cout<<"100"<<endl;
    return 0;
}

```