//This solution is not worked out by myself.
#include<cstdio>
long long str[205][2];
long long n,noww=100;
void solve(long long num)
{
	if(num<=1) return ;
	if(num%2) str[++str[0][0]][0]=noww--,solve(num-1);
	else str[++str[0][1]][1]=noww--,solve(num/2);
}
int main ()
{
	scanf("%lld",&n); solve(n+1);
	printf("%lld\n",2*(str[0][0]+str[0][1])); noww=100-str[0][0]-str[0][1]+1;
	for(int i=1;i<=str[0][0];i++) printf("%lld ",str[i][0]);
	for(int i=str[0][1];i;i--) printf("%lld ",str[i][1]);
	for(int i=noww;i<=100;i++) printf("%lld ",1ll*i);
	return 0;
}
### Prompt

Please provide a CPP coded solution 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
//This solution is not worked out by myself.
#include<cstdio>
long long str[205][2];
long long n,noww=100;
void solve(long long num)
{
	if(num<=1) return ;
	if(num%2) str[++str[0][0]][0]=noww--,solve(num-1);
	else str[++str[0][1]][1]=noww--,solve(num/2);
}
int main ()
{
	scanf("%lld",&n); solve(n+1);
	printf("%lld\n",2*(str[0][0]+str[0][1])); noww=100-str[0][0]-str[0][1]+1;
	for(int i=1;i<=str[0][0];i++) printf("%lld ",str[i][0]);
	for(int i=str[0][1];i;i--) printf("%lld ",str[i][1]);
	for(int i=noww;i<=100;i++) printf("%lld ",1ll*i);
	return 0;
}
```