#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
#define ll long long
ll n;int cnt;
deque<int> Q1,Q2;
void Work(ll n)
{
	if(n==1)return;
	Work(n>>1);++cnt;
	Q1.push_front(cnt);Q2.push_front(cnt);
	if(n&1)++cnt,Q1.push_front(cnt),Q2.push_back(cnt); 
}
int main()
{
	cin>>n;Work(n+1);
	printf("%d\n",(int)(Q1.size()+Q2.size()));
	while(!Q1.empty())printf("%d ",Q1.front()),Q1.pop_front();
	while(!Q2.empty())printf("%d ",Q2.front()),Q2.pop_front();
	puts("");return 0;
}

### Prompt

Create a solution in Cpp for 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<iostream>
#include<cstdio>
#include<queue>
using namespace std;
#define ll long long
ll n;int cnt;
deque<int> Q1,Q2;
void Work(ll n)
{
	if(n==1)return;
	Work(n>>1);++cnt;
	Q1.push_front(cnt);Q2.push_front(cnt);
	if(n&1)++cnt,Q1.push_front(cnt),Q2.push_back(cnt); 
}
int main()
{
	cin>>n;Work(n+1);
	printf("%d\n",(int)(Q1.size()+Q2.size()));
	while(!Q1.empty())printf("%d ",Q1.front()),Q1.pop_front();
	while(!Q2.empty())printf("%d ",Q2.front()),Q2.pop_front();
	puts("");return 0;
}

```