//Created Time:2019年12月18日 星期三 18时32分15秒
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

long long n;
int a[2002];
int cnt;

int main(){
	cin >> n; ++n;
	int l = 0, r = 50; bool flag = false;
	for(int i = 39, pos = 1; ~i; --i){
		long long p = 1ll << i;
		if(p > n) continue;
		if(p & n){
			if(flag)
				++cnt, a[++l] = pos++;
			else flag = true;
		}
		if(i) a[++r] = pos++, ++cnt;
	}
	cout << cnt + 100 << endl;
	for(int i = l; i; --i) cout << a[i] << ' ';
	for(int i = 51; i <= r; ++i) cout << a[i] << ' ';
	for(int i = 1; i <= 100; ++i) cout << i << ' ';
	return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

Examples

Input

7


Output

4
1 1 1 1


Input

299


Output

23
32 11 11 73 45 8 11 83 83 8 45 32 32 10 100 73 32 83 45 73 32 11 10

### Response

```cpp
//Created Time:2019年12月18日 星期三 18时32分15秒
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

long long n;
int a[2002];
int cnt;

int main(){
	cin >> n; ++n;
	int l = 0, r = 50; bool flag = false;
	for(int i = 39, pos = 1; ~i; --i){
		long long p = 1ll << i;
		if(p > n) continue;
		if(p & n){
			if(flag)
				++cnt, a[++l] = pos++;
			else flag = true;
		}
		if(i) a[++r] = pos++, ++cnt;
	}
	cout << cnt + 100 << endl;
	for(int i = l; i; --i) cout << a[i] << ' ';
	for(int i = 51; i <= r; ++i) cout << a[i] << ' ';
	for(int i = 1; i <= 100; ++i) cout << i << ' ';
	return 0;
}

```