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 <bits/stdc++.h>

using namespace std;

#define int long long
#define FOR(i, a, b) for(ll i = (a); i < (b); ++i)
#define FORR(i, a, b) for(ll i = (a); i > (b); --i)
#define REP(i, n) for(ll i = 0; i < (n); ++i)
#define REPR(i, n) for(ll i = n; i >= 0; i--)
#define FOREACH(x, a) for(auto &(x) : (a))
#define VECCIN(x)                                                              \
    for(auto &youso_ : (x)) cin >> youso_
#define bitcnt __builtin_popcount
#define SZ(x) ((ll)(x).size())
#define fi first
#define se second
#define All(a) (a).begin(), (a).end()
#define rAll(a) (a).rbegin(), (a).rend()
template <typename T = long long> inline T IN() {
    T x;
    cin >> x;
    return (x);
}
inline void CIN() {}
template <class Head, class... Tail>
inline void CIN(Head &&head, Tail &&... tail) {
    cin >> head;
    CIN(move(tail)...);
}
#define CCIN(...)                                                              \
    char __VA_ARGS__;                                                          \
    CIN(__VA_ARGS__)
#define DCIN(...)                                                              \
    double __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define LCIN(...)                                                              \
    ll __VA_ARGS__;                                                            \
    CIN(__VA_ARGS__)
#define SCIN(...)                                                              \
    string __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define Yes(a) cout << (a ? "Yes" : "No") << "\n"
#define YES(a) cout << (a ? "YES" : "NO") << "\n"
#define Printv(v)                                                              \
    {                                                                          \
        FOREACH(x, v) { cout << x << " "; }                                    \
        cout << "\n";                                                          \
    }
template <typename T = string> inline void eputs(T s) {
    cout << s << "\n";
    exit(0);
}
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    std::fill((T *)array, (T *)(array + N), val);
}
template <typename T> using PQG = priority_queue<T, vector<T>, greater<T>>;
template <typename T> using PQ = priority_queue<T>;

typedef long long ll;
typedef pair<ll, ll> PL;
typedef vector<PL> VPL;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef vector<double> VD;

const int INF = 1e9;
// const int MOD = 1e9 + 7;
const int MOD = 998244353;
const ll LINF = 1e18;
const double PI = atan(1.0) * 4.0;
const ll dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
#define PI 3.141592653589793238

ll N, M;
VVL G, ans;
vector<bool> used(100100, false);
map<PL, ll> eidx;

ll dfs(ll now, ll par) {
    ll p1 = -1;
    FOREACH(to, G[now]) {
        ll idx = eidx[{now, to}];
        if(used[idx]) continue;
        used[idx] = true;
        ll p2 = dfs(to, now);
        if(p2 != -1) {
            if(p1 == -1)
                p1 = p2;
            else {
                ans[now].emplace_back(p1);
                ans[now].emplace_back(p2);
                p1 = -1;
            }
        }
    }
    if(p1 != -1) {
        ans[now].emplace_back(p1);
        ans[now].emplace_back(par);
        return -1;
    }
    return now;
}

signed main() {
    LCIN(N);
    ll ma = 60;
    for(; ma; ma--) {
        if(N >= 1LL << ma) break;
    }
    VL ans;
    REP(i, ma) ans.emplace_back(i + 1);
    N -= 1LL << ma;
    ll next = ma + 1;
    REPR(i, ma - 1) {
        if(N >= 1LL << i) {
            ans.emplace(ans.begin() + i, next);
            next++;
            N -= 1LL << i;
        }
    }
    REP(i, next) ans.emplace_back(i + 1);
    cout << ans.size() << "\n";
    REP(i, ans.size()) cout << ans[i] << (i == ans.size() - 1 ? "\n" : " ");
}