#include <stdio.h>
#include <iostream> 
using namespace std;
const int maxn=105;
int odd[maxn],even[maxn];
long long n;
int main(){
    int i,num=1;
    scanf("%lld",&n);n++;
    while(n>1){
        if(n&1)odd[++odd[0]]=num++,n--;
        else even[++even[0]]=num++,n/=2;
    }
    printf("%d\n",odd[0]+even[0]<<1);
    for(i=1;i<=odd[0];i++)printf("%d ",odd[i]);
    for(i=even[0];i>=1;i--)printf("%d ",even[i]);
    for(i=odd[0]+even[0];i>=1;i--)printf("%d ",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 <stdio.h>
#include <iostream> 
using namespace std;
const int maxn=105;
int odd[maxn],even[maxn];
long long n;
int main(){
    int i,num=1;
    scanf("%lld",&n);n++;
    while(n>1){
        if(n&1)odd[++odd[0]]=num++,n--;
        else even[++even[0]]=num++,n/=2;
    }
    printf("%d\n",odd[0]+even[0]<<1);
    for(i=1;i<=odd[0];i++)printf("%d ",odd[i]);
    for(i=even[0];i>=1;i--)printf("%d ",even[i]);
    for(i=odd[0]+even[0];i>=1;i--)printf("%d ",i);
    return 0;
}
```