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 FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)

#define INF INT_MAX/3
#define ALL(a) (a).begin(),(a).end()
#define SET(a,c) memset(a,c,sizeof a)
#define CLR(a) memset(a,0,sizeof a)
#define ll long long
#define ull unsigned long long

#define PI (3.1415926535897932)
#define eps 1e-8

void solve(ll N) {
	int cnt = -2; int fwd[101] = {};

	for (ll i = N; i > 0; i >>= 1) {
		cnt++;
		if (i & 1) {
			cnt++;
			fwd[cnt] = 1;
		}
	}

	cout << cnt + 100 << endl;
	FOR(i, 1, 101)cout << i << " ";
	RFOR(i, 0, cnt) {
		if (fwd[cnt - i - 1] == 1)cout << i + 1 << " ";
	}
	FOR(i, 0, cnt) {
		if (fwd[cnt - i - 1] == 0)cout << i + 1 << " ";
	}
	cout << endl;
}

int main() {
	ll N;
	cin >> N;
	solve(N);
}