#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn=2010;
ll n;
int ans[maxn], cnt, a[maxn];

vector<int> g;

int main(){

	scanf("%lld", &n);
	int tot=0;
	for(int i=50;i>=0;i--){
		if((1ll<<i)-1<=n){ tot=i; break; }
	}
	// printf("%lld\n", 1ll<<tot);
	n-=(1ll<<tot)-1;
	for(int i=1;i<=tot;i++) g.push_back(i);
	int m=tot;
	// printf("n = %lld\n", n);
	for(int i=(tot-1);i>=0;i--){
		if(!(n & (1ll<<i))) continue;
		m++;
		// g.insert(m,g.begin()+i);
		g.insert(g.begin()+i, m);
	}
	for(int i=1;i<=m;i++) g.push_back(i);
	printf("%d\n", 2*m);
	for(int i=0;i<m*2;i++) printf("%d ", g[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
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn=2010;
ll n;
int ans[maxn], cnt, a[maxn];

vector<int> g;

int main(){

	scanf("%lld", &n);
	int tot=0;
	for(int i=50;i>=0;i--){
		if((1ll<<i)-1<=n){ tot=i; break; }
	}
	// printf("%lld\n", 1ll<<tot);
	n-=(1ll<<tot)-1;
	for(int i=1;i<=tot;i++) g.push_back(i);
	int m=tot;
	// printf("n = %lld\n", n);
	for(int i=(tot-1);i>=0;i--){
		if(!(n & (1ll<<i))) continue;
		m++;
		// g.insert(m,g.begin()+i);
		g.insert(g.begin()+i, m);
	}
	for(int i=1;i<=m;i++) g.push_back(i);
	printf("%d\n", 2*m);
	for(int i=0;i<m*2;i++) printf("%d ", g[i]);
	return 0;
}
```