Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

7

Sample Output 1

4
1 1 1 1

There are two good strings that appear as subsequences of s: (1,1) and (1,1,1,1). There are six occurrences of (1,1) and one occurrence of (1,1,1,1), for a total of seven.

Sample Input 2

299

Sample Output 2

23
32 11 11 73 45 8 11 83 83 8 45 32 32 10 100 73 32 83 45 73 32 11 10

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

    #include <bits/stdc++.h>
     
    #define fr first
    #define sc second
    #define mk make_pair
    #define pb push_back
    #define ll long long
    #define ld long double
    #define OK puts("OK");
    #define sz(s) (int)s.size()
    #define all(s) s.begin(), s.end()
     
    using namespace std;
     
    const int N = 2e5 + 10;
    const int inf = 1e9 + 7;
     
    ll n, cnt;
    deque <int> dq;
     
    void go(ll a)
    {
        if(a == 1)
            return;
        if(a & 1)
        {
            go(a - 1);
            dq.push_front(++ cnt);
            return;
        }
        go(a / 2);
        dq.pb(++ cnt);
    }
     
    main()
    {
        cin >> n;
        go(n + 1);
        cout << 2 * cnt << endl;
        while(!dq.empty())
            cout << dq.front() << " ", dq.pop_front();
        for(int i = 1 ; i <= cnt ; i ++)
            cout << i << " ";
    }