#include <bits/stdc++.h>

using namespace std;

int a[500], nr = 0;
long long n;

void get_ans(long long k, int id1, int id2)
{
    if(k==1)
    {
        int i;
        cout << 2*nr << '\n';
        for(i=1; i<id1; ++i) cout << a[i] << ' ';
        for(i=id2+1; i<=100; ++i) cout << a[i] << ' ';

        for(i=nr; i; --i) cout << i << ' ';
        cout << '\n';
        return;
    }

    if(k&1LL)
    {
        a[id1] = ++nr;
        get_ans(k-1, id1+1, id2);
    }
    else
    {
        a[id2] = ++nr;
        get_ans(k/2, id1, id2-1);
    }
}

int main()
{
    cin >> n; nr = 0;
    get_ans(n+1, 1, 100);

    return 0;
}

### Prompt

Your challenge is to write 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 <bits/stdc++.h>

using namespace std;

int a[500], nr = 0;
long long n;

void get_ans(long long k, int id1, int id2)
{
    if(k==1)
    {
        int i;
        cout << 2*nr << '\n';
        for(i=1; i<id1; ++i) cout << a[i] << ' ';
        for(i=id2+1; i<=100; ++i) cout << a[i] << ' ';

        for(i=nr; i; --i) cout << i << ' ';
        cout << '\n';
        return;
    }

    if(k&1LL)
    {
        a[id1] = ++nr;
        get_ans(k-1, id1+1, id2);
    }
    else
    {
        a[id2] = ++nr;
        get_ans(k/2, id1, id2-1);
    }
}

int main()
{
    cin >> n; nr = 0;
    get_ans(n+1, 1, 100);

    return 0;
}

```