#include <cstdio>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
#define repd(i,x,y) for (int i=x; i>=y; --i)
using namespace std;

typedef long long LL;
int head,tail,q[210],m,a[110];
LL n;

int main()
{
    scanf("%lld",&n);
    head=100,tail=99;
    while (n)
        if (n&1)
            n>>=1,a[++m]=1;
        else
            --n,++m;
    repd(i,m,1)
        (a[i]?q[++tail]:q[--head])=m-i+1;
    printf("%d\n",m*2);
    repu(i,head,tail)
        printf("%d ",q[i]);
    repu(i,1,m)
        printf("%d ",i);
    return 0;
}
### Prompt

Create a solution in CPP for 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>
#define repu(i,x,y) for (int i=x; i<=y; ++i)
#define repd(i,x,y) for (int i=x; i>=y; --i)
using namespace std;

typedef long long LL;
int head,tail,q[210],m,a[110];
LL n;

int main()
{
    scanf("%lld",&n);
    head=100,tail=99;
    while (n)
        if (n&1)
            n>>=1,a[++m]=1;
        else
            --n,++m;
    repd(i,m,1)
        (a[i]?q[++tail]:q[--head])=m-i+1;
    printf("%d\n",m*2);
    repu(i,head,tail)
        printf("%d ",q[i]);
    repu(i,1,m)
        printf("%d ",i);
    return 0;
}
```