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)

//              +-- -- --++-- +-In the name of ALLAH-+ --++-- -- --+              \\

#include <bits/stdc++.h>

#define int ll

#define F first
#define S second
#define _sz(x) (int)x.size()
#define pb push_back

#define error(x) cout << #x << " = " << x << endl

using namespace std ;
using ll = long long ;
using ld = long double ;
using pii = pair <int , int> ;

const int C = 100 + 10;
int n, a[C], dp[C], sum;

int32_t main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

	cin >> n;

	if (n <= 100)
	{
		for (int i = 0; i < n; i++) cout << i + 1 << ' ';
		for (int i = 0; i < n; i++) cout << i + 1 << ' ';
		cout << '\n';
		return 0;
	}


	sum = 100;

	for (int i = 0; i < 100; i ++)
	{
		dp[i] = 1;

		vector <pii> vec;

		for (int j = 0; j < i; j++) vec.pb({a[j], dp[j]});

		sort(vec.begin(), vec.end());

		int p = 1;
		for (int j = 0; j < i; j++)
		{
			if (sum + vec[j].S > n) break;
			p++;
			dp[i] += vec[j].S;
		}

		sum += dp[i] - 1;

		a[i] = p;
		for (int j = 0; j < i; j++)
			if (a[j] >= p) a[j]++;
	}

	for (int i = 0; i < 100; i++) cout << i + 1 << ' ';
	for (int i = 0; i < 100; i++) cout << a[i] << ' ';
	cout << '\n';
}