//2018-1-20
//miaomiao
//
#include <bits/stdc++.h>
using namespace std;

#define LL long long
#define For(i, a, b) for(int i = (a); i <= (int)(b); ++i)

#define N (200+5)
const int A = 100;

int m, ans[N], l, r;

void Solve(LL n){
	if(n <= 1) return;

	if(n & 1){
		Solve(n-1);
		ans[--l] = ++m;
	}else{
		Solve(n>>1);
		ans[++r] = ++m;
	}
}

int main(){
	LL n;

	scanf("%lld", &n); ++n;

	l = A+1; r = A;
	Solve(n);
	
	printf("%d\n", m<<1);
	For(i, l, r) printf("%d ", ans[i]);
	For(i, 1, m) printf("%d%c", i, i==m? '\n': ' ');

	return 0;
}
### Prompt

Your challenge is to write a CPP solution 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
//2018-1-20
//miaomiao
//
#include <bits/stdc++.h>
using namespace std;

#define LL long long
#define For(i, a, b) for(int i = (a); i <= (int)(b); ++i)

#define N (200+5)
const int A = 100;

int m, ans[N], l, r;

void Solve(LL n){
	if(n <= 1) return;

	if(n & 1){
		Solve(n-1);
		ans[--l] = ++m;
	}else{
		Solve(n>>1);
		ans[++r] = ++m;
	}
}

int main(){
	LL n;

	scanf("%lld", &n); ++n;

	l = A+1; r = A;
	Solve(n);
	
	printf("%d\n", m<<1);
	For(i, l, r) printf("%d ", ans[i]);
	For(i, 1, m) printf("%d%c", i, i==m? '\n': ' ');

	return 0;
}
```