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: Wrong Answer)

#include <algorithm>
#include <cmath>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rrep(i, n) for (int i = (n)-1; i >= 0; i--)
using namespace std;
using lli = long long int;
lli val[40];
lli memo[100][100] = {};
lli cmb(lli n, lli r)
{
    if (memo[n][r])
        return memo[n][r];
    if (n == 1)
        return 1;
    if (r == 0 || r == n)
        return 1;
    else
        return memo[n][r] = cmb(n - 1, r - 1) + cmb(n - 1, r);
}
int main()
{


    /*
     * 同じものを2*k回繰り返したものは部分列として
     * 2kC2 + 2kC4 + 2kC6 ... + のよいあれを持つ(k==1)の場合奇数を作れる
     */
    lli n;
    cin >> n;
    for (int i = 1; i < 25; i++) {
        for (int j = 2; j <= 2 * i; j += 2) {
            val[i] += cmb(2 * i, j);
        }
    }
    //cout << val[24] << endl;
    int cnt = 1;
    //cout << val[1] << endl;
    vector<int> ans;
    vector<int> last;
    bool flag = false;
    for (int i = 24; i > 0; i--) {
        if (n >= val[i]) {
            lli g = n / val[i];
            if (!flag) {
                rep(k, 2 * i)
                {
                    ans.push_back(cnt);
                }
                cnt++;
                flag = true;
                n -= val[i];
            } else {
                int a = 0;
                rep(k, g)
                {
                    rep(j, ans.size())
                    {
                        if (ans[j] == 1)
                            a++;
                        if (a == i + 1) {
                            ans.insert(ans.begin() + j, cnt);
                            last.push_back(cnt);
                            cnt++;
                        }
                    }
                }
                n -= val[i] * g;
            }
        }
    }
    reverse(last.begin(), last.end());
    cout << ans.size() + last.size() << endl;
    for (auto s : ans) {
        cout << s << " ";
    }
    for (auto s : last) {
        cout << s << " ";
    }
    cout << endl;
}