#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n;
int i,j,k,m,q[211],a[211];
bool cmp(int x,int y){
	return q[x]<q[y];
}
int main(){
	scanf("%lld",&n);n++;
	for (m=1;(1LL<<m)<=n;m++) q[m]=2*m;
	m--;n-=(1LL<<m);
	while (n){
		for (i=0;(1LL<<i)<=n;i++);
		i--;n-=(1LL<<i);q[++m]=2*i+1;
	}
	for (i=1;i<=m;i++) a[i]=i;
	sort(a+1,a+1+m,cmp);
	printf("%d\n",2*m);
	for (i=1;i<=m;i++) printf("%d ",a[i]);
	for (i=1;i<=m;i++) printf("%d ",i);
	puts("");
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n;
int i,j,k,m,q[211],a[211];
bool cmp(int x,int y){
	return q[x]<q[y];
}
int main(){
	scanf("%lld",&n);n++;
	for (m=1;(1LL<<m)<=n;m++) q[m]=2*m;
	m--;n-=(1LL<<m);
	while (n){
		for (i=0;(1LL<<i)<=n;i++);
		i--;n-=(1LL<<i);q[++m]=2*i+1;
	}
	for (i=1;i<=m;i++) a[i]=i;
	sort(a+1,a+1+m,cmp);
	printf("%d\n",2*m);
	for (i=1;i<=m;i++) printf("%d ",a[i]);
	for (i=1;i<=m;i++) printf("%d ",i);
	puts("");
	return 0;
}
```