#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
list <int> ans;
bool vis[50] = {0};
int main()
{
    LL n;
    scanf("%lld" , &n);
    ++n;
    int p = 0 , q = 0;
    while(n)
    {
        vis[p++] = n & 1;
        n >>= 1;
    }
    for(int i = p - 2 ; i >= 0 ; --i)
    {
        ans.push_back(++q);
        if(vis[i])
            ans.push_front(++q);
    }
    printf("%d\n" , q + q);
    for(int i = 1 ; i <= q ; ++i)
        printf("%d " , i);
    for(int i = 1 ; i <= q ; ++i)
    {
        printf("%d" , ans.front());
        putchar(" \n"[i == q + 1]);
        ans.pop_front();
    }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 <bits/stdc++.h>
using namespace std;
typedef long long LL;
list <int> ans;
bool vis[50] = {0};
int main()
{
    LL n;
    scanf("%lld" , &n);
    ++n;
    int p = 0 , q = 0;
    while(n)
    {
        vis[p++] = n & 1;
        n >>= 1;
    }
    for(int i = p - 2 ; i >= 0 ; --i)
    {
        ans.push_back(++q);
        if(vis[i])
            ans.push_front(++q);
    }
    printf("%d\n" , q + q);
    for(int i = 1 ; i <= q ; ++i)
        printf("%d " , i);
    for(int i = 1 ; i <= q ; ++i)
    {
        printf("%d" , ans.front());
        putchar(" \n"[i == q + 1]);
        ans.pop_front();
    }
}

```