#include<iostream>
using namespace std;
long long n,now=0,good,targ=0,log2=0;
int a[205]={0};
void solution(long long good)
{
    int i=40, L=101,R=L-1;
    for(;!((1ll<<i)&good);i--);
    for(i--;i>=0;i--){
        a[++R]=++now;
        if((1ll<<i)&good) a[--L]=++now;
    }

    n=R-L+1;
    for(int i=1;i<=n;i++) a[i]=a[i+L-1];
    for(int i=n+1;i<=n*2;i++) a[i]=i-n;
    n<<=1;
}
int main()
{
    cin>>good;
    int tep=good+1;
    while(tep/2){
        log2++;
        tep/=2;
    }
    targ=log2;
    solution(good+1);
    cout<<n<<endl;
    for(int i=1;i<=n;i++){
        cout<<a[i]<<" ";
    }
}

### Prompt

Please provide a CPP coded solution 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>
using namespace std;
long long n,now=0,good,targ=0,log2=0;
int a[205]={0};
void solution(long long good)
{
    int i=40, L=101,R=L-1;
    for(;!((1ll<<i)&good);i--);
    for(i--;i>=0;i--){
        a[++R]=++now;
        if((1ll<<i)&good) a[--L]=++now;
    }

    n=R-L+1;
    for(int i=1;i<=n;i++) a[i]=a[i+L-1];
    for(int i=n+1;i<=n*2;i++) a[i]=i-n;
    n<<=1;
}
int main()
{
    cin>>good;
    int tep=good+1;
    while(tep/2){
        log2++;
        tep/=2;
    }
    targ=log2;
    solution(good+1);
    cout<<n<<endl;
    for(int i=1;i<=n;i++){
        cout<<a[i]<<" ";
    }
}

```