#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)

typedef long long LL;
const int N=20005;

int tot;

std:: deque <int> L,R;

void solve(LL n) {
	if (n==1) return ;
	solve(n/2);
	L.push_front(++tot);
	R.push_front(tot);
	if (n&1) {
		L.push_front(++tot);
		R.push_back(tot);
	}
}

int main(void) {
	LL n; scanf("%lld",&n);
	solve(n+1);
	printf("%d\n", L.size()+R.size());
	for (int i=0;i<L.size();++i) printf("%d ", L[i]);
	for (int i=0;i<R.size();++i) printf("%d ", R[i]);
	puts("");
	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 <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)

typedef long long LL;
const int N=20005;

int tot;

std:: deque <int> L,R;

void solve(LL n) {
	if (n==1) return ;
	solve(n/2);
	L.push_front(++tot);
	R.push_front(tot);
	if (n&1) {
		L.push_front(++tot);
		R.push_back(tot);
	}
}

int main(void) {
	LL n; scanf("%lld",&n);
	solve(n+1);
	printf("%d\n", L.size()+R.size());
	for (int i=0;i<L.size();++i) printf("%d ", L[i]);
	for (int i=0;i<R.size();++i) printf("%d ", R[i]);
	puts("");
	return 0;
}
```