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;
typedef long long ll;

vector<int> construct(ll N) {
    int n;
    vector<int> p;
    for(n=40; n>0; n--) {
        if(N >= (1LL<<n)-1) break;
    }
    for(int i=1; i<=n; i++) p.emplace_back(i);
    N -= (1LL<<n) - 1;

    for(int i=n-1; i>=0; i--) {
        if(N < 1LL<<i) continue;
        p.insert(p.begin()+i, n++);
        N -= 1LL<<i;
    }

    return p;
}

int main(void) {
    ll N, tmp;
    int num = 0;
    cin >> N;
    tmp = N + 1;
    vector<int> ans;

    while(tmp) {tmp /= 2; num++;}
    for(int i=0; i<num-1; i++) {
        if(((N+1) >> i) & 1) ans.emplace_back(100-i);
        ans.emplace_back(i+1);
    }

    printf("%d\n", ans.size())+100;
    for(int a : ans) printf("%d ", a);
    for(int i=1; i<=100; i++) printf("%d ", i);
    printf("\n");
    return 0;
}