# include <stdio.h>
# include <iostream>
using namespace std;
int a[103], b[103];

int main()

{

    int p1=0, p2=0, m=100;

    long long n;

    scanf("%lld",&n);

    n++;

    while(n > 1)

    {

        if(n&1) a[++p1] = m--, --n;

        else b[++p2] = m--, n>>=1;

    }

    printf("%d\n",p1+p2<<1);

    for(int i=1; i<=p1; ++i)

        printf("%d ",a[i]);

    for(int i=p2; i>=1; --i)

        printf("%d ",b[i]);


    for(int i=101-p1-p2; i<=100; ++i)

        printf("%d ",i);

    puts("");

    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 <stdio.h>
# include <iostream>
using namespace std;
int a[103], b[103];

int main()

{

    int p1=0, p2=0, m=100;

    long long n;

    scanf("%lld",&n);

    n++;

    while(n > 1)

    {

        if(n&1) a[++p1] = m--, --n;

        else b[++p2] = m--, n>>=1;

    }

    printf("%d\n",p1+p2<<1);

    for(int i=1; i<=p1; ++i)

        printf("%d ",a[i]);

    for(int i=p2; i>=1; --i)

        printf("%d ",b[i]);


    for(int i=101-p1-p2; i<=100; ++i)

        printf("%d ",i);

    puts("");

    return 0;

}
```