#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
using namespace std ;

int d[500], opt[500] ;
ll n ;

int main() {
	scanf("%lld", &n) ;
	int k = 0 ; ll tmp = n + 1 ;
	for ( ; tmp ; ++ k) if (tmp & 1LL) -- tmp, d[k] = 0 ; else tmp /= 2, d[k] = 1 ;
	int he = k, ta = he - 1 ;
	int top = 0 ;
	for (int i = k - 2 ; i >= 0; -- i) if (d[i]) opt[++ ta] = ++ top ; else opt[-- he] = ++ top ;
	printf("%d\n", k * 2 - 2) ;
	rep(i, he, ta) printf("%d ", opt[i]) ;
	rep(i, 1, k - 1) printf("%d ", i) ;

	return 0 ; 
}
### Prompt

Your task is to create 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
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
using namespace std ;

int d[500], opt[500] ;
ll n ;

int main() {
	scanf("%lld", &n) ;
	int k = 0 ; ll tmp = n + 1 ;
	for ( ; tmp ; ++ k) if (tmp & 1LL) -- tmp, d[k] = 0 ; else tmp /= 2, d[k] = 1 ;
	int he = k, ta = he - 1 ;
	int top = 0 ;
	for (int i = k - 2 ; i >= 0; -- i) if (d[i]) opt[++ ta] = ++ top ; else opt[-- he] = ++ top ;
	printf("%d\n", k * 2 - 2) ;
	rep(i, he, ta) printf("%d ", opt[i]) ;
	rep(i, 1, k - 1) printf("%d ", i) ;

	return 0 ; 
}
```