#include<bits/stdc++.h>
using namespace std;
#define RI register int
typedef long long LL;
LL bin[42],n;
int a[205],js,len,k;
int main()
{
	bin[0]=1;for(RI i=1;i<=40;++i) bin[i]=bin[i-1]<<1LL;
	scanf("%lld",&n);++n;
	for(RI i=40;i>=0;--i) if(n&bin[i]) {k=i;break;}
	for(RI i=k-1;i>=0;--i) {
		++js;
		for(RI j=len;j>=len/2+1;--j) a[j+2]=a[j];
		a[len/2+2]=js;
		for(RI j=len/2;j>=1;--j) a[j+1]=a[j];
		a[1]=js,len+=2;
		if(n&bin[i]) {
			++js;
			for(RI j=len;j>=1;--j) a[j+1]=a[j];
			a[1]=a[len+2]=js,len+=2;
		}
	}
	printf("%d\n",len);
	for(RI i=1;i<=len;++i) printf("%d ",a[i]);
	puts("");
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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;
#define RI register int
typedef long long LL;
LL bin[42],n;
int a[205],js,len,k;
int main()
{
	bin[0]=1;for(RI i=1;i<=40;++i) bin[i]=bin[i-1]<<1LL;
	scanf("%lld",&n);++n;
	for(RI i=40;i>=0;--i) if(n&bin[i]) {k=i;break;}
	for(RI i=k-1;i>=0;--i) {
		++js;
		for(RI j=len;j>=len/2+1;--j) a[j+2]=a[j];
		a[len/2+2]=js;
		for(RI j=len/2;j>=1;--j) a[j+1]=a[j];
		a[1]=js,len+=2;
		if(n&bin[i]) {
			++js;
			for(RI j=len;j>=1;--j) a[j+1]=a[j];
			a[1]=a[len+2]=js,len+=2;
		}
	}
	printf("%d\n",len);
	for(RI i=1;i<=len;++i) printf("%d ",a[i]);
	puts("");
    return 0;
}
```