#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
stack<bool> stk;
vector<int> v1, v2;
#define pb push_back
int main(){
	ll n; cin >> n;
	n = n + 1;
	ll tmp = n;
	for(;tmp; tmp >>= 1) stk.push(tmp & 1);
	stk.pop();
	for(int x = 1; !stk.empty(); stk.pop()){
		v1.pb(x), ++x;
		if(stk.top()) v2.pb(x), ++x;
	}
	int m = v1.size() + v2.size();
	cout << m * 2 << endl;
	for(int i = 1; i <= m; ++i)
		cout << i << " ";
	for(int i = v2.size() - 1; ~i; --i)
		cout << v2[i] << " ";
	for(auto x: v1)
		cout << x << " ";
	puts("");
	return 0;
}
### Prompt

Your challenge is to write 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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
stack<bool> stk;
vector<int> v1, v2;
#define pb push_back
int main(){
	ll n; cin >> n;
	n = n + 1;
	ll tmp = n;
	for(;tmp; tmp >>= 1) stk.push(tmp & 1);
	stk.pop();
	for(int x = 1; !stk.empty(); stk.pop()){
		v1.pb(x), ++x;
		if(stk.top()) v2.pb(x), ++x;
	}
	int m = v1.size() + v2.size();
	cout << m * 2 << endl;
	for(int i = 1; i <= m; ++i)
		cout << i << " ";
	for(int i = v2.size() - 1; ~i; --i)
		cout << v2[i] << " ";
	for(auto x: v1)
		cout << x << " ";
	puts("");
	return 0;
}
```