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>
#define DEBUG 0
/*
*/

using namespace std;

uint64_t combinations2(uint64_t n, uint64_t k) {
	if (n < k)
	{
		return 0;
	}
	uint64_t r = 1;
	for (uint64_t d = 1; d <= k; ++d) {
		r *= n--;
		r /= d;
	}
	return r;
}

int main() {
	long long int N;
	cin >> N;
	vector<long long int>ans;
	long long int now = 1;
	while (N > 0)
	{
		long long int M = 0;
		for (size_t i = 1; i < 60; i++)
		{
			long long int sum = 0;
			for (size_t j = 2; j <= i; j += 2)
			{
				sum += combinations2(i, j);
			}
			//cout << sum << " " << N << endl;
			if (sum > N)
			{
				for (size_t j = 0; j < i; j++)
				{
					ans.push_back(now);
				}
				now++;
				N -= M;
				M = 0;
				break;
			}
			else {
				M = sum;
			}
		}
	}
	cout << ans.size() << endl;
	for (size_t i = 0; i < ans.size() - 1; i++)
	{
		cout << ans[i] << " ";
	}
	cout << ans[ans.size() - 1] << endl;


	if (DEBUG)
	{
		int ewataeoprgjpsrteog;
		cin >> ewataeoprgjpsrteog;
	}
}