#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<deque>
#define fo(i,j,k) for(int i=j;i<=k;++i)
#define fd(i,j,k) for(int i=j;i>=k;--i)
#define ll long long
using namespace std;
const int N=50;
int a[N],tot=0;
deque<int> q;
void dfs(ll x){
	if(x==1) return;
	if(x&1) dfs(x-1),q.push_front(++tot);
	else dfs(x>>1),q.push_back(++tot);
}
int main()
{
	ll n;
	scanf("%lld",&n);
	dfs(n+1),printf("%d\n",tot<<1);
	fo(i,0,tot-1) printf("%d ",q[i]);
	fo(i,1,tot) printf("%d ",i);
}

### Prompt

Please create a solution in cpp 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<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<deque>
#define fo(i,j,k) for(int i=j;i<=k;++i)
#define fd(i,j,k) for(int i=j;i>=k;--i)
#define ll long long
using namespace std;
const int N=50;
int a[N],tot=0;
deque<int> q;
void dfs(ll x){
	if(x==1) return;
	if(x&1) dfs(x-1),q.push_front(++tot);
	else dfs(x>>1),q.push_back(++tot);
}
int main()
{
	ll n;
	scanf("%lld",&n);
	dfs(n+1),printf("%d\n",tot<<1);
	fo(i,0,tot-1) printf("%d ",q[i]);
	fo(i,1,tot) printf("%d ",i);
}

```