#include <cstdio>
#include <cstdlib>
#include <iostream>
using namespace std;
typedef long long LL;
const int maxn=110;
int ans[maxn*2];
int main()
{
    LL n;
    cin>>n;
    n++;
    int tot=0;
    int w=0;
    while((n>>w)>1)w++;
    int cnt=0;
    while(w>0)
    {
        ans[++tot]=++cnt;
        w--;
        if((n>>w)&1)
        {
            ++tot;
            for(int i=tot;i>1;i--) ans[i]=ans[i-1];
            ans[1]=++cnt;
        }
    }
    printf("%d\n",cnt*2);
    for(int i=1;i<=cnt;i++) printf("%d ",i);
    for(int i=1;i<=cnt;i++) printf("%d ",ans[i]);
    return 0;
}

### Prompt

Generate 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 <cstdio>
#include <cstdlib>
#include <iostream>
using namespace std;
typedef long long LL;
const int maxn=110;
int ans[maxn*2];
int main()
{
    LL n;
    cin>>n;
    n++;
    int tot=0;
    int w=0;
    while((n>>w)>1)w++;
    int cnt=0;
    while(w>0)
    {
        ans[++tot]=++cnt;
        w--;
        if((n>>w)&1)
        {
            ++tot;
            for(int i=tot;i>1;i--) ans[i]=ans[i-1];
            ans[1]=++cnt;
        }
    }
    printf("%d\n",cnt*2);
    for(int i=1;i<=cnt;i++) printf("%d ",i);
    for(int i=1;i<=cnt;i++) printf("%d ",ans[i]);
    return 0;
}

```