#include <cstdio>
#include <deque>
using namespace std;

typedef long long i64;

i64 n;
int sz;
deque<int> a, b;

void build(i64 n) {
	if (n == 0) return;

	if (n % 2 == 1) {
		build(n / 2);
		++sz;
		a.push_back(sz);
		b.push_back(sz);
	} else {
		build(n - 1);
		++sz;
		a.push_back(sz);
		b.push_front(sz);
	}
}

int main() {
	scanf("%lld", &n);
	build(n);
	printf("%d\n", 2 * (int) a.size());

	for (int i : a) printf("%d ", i);
	for (int i : b) printf("%d ", i);
	puts("");
	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
#include <cstdio>
#include <deque>
using namespace std;

typedef long long i64;

i64 n;
int sz;
deque<int> a, b;

void build(i64 n) {
	if (n == 0) return;

	if (n % 2 == 1) {
		build(n / 2);
		++sz;
		a.push_back(sz);
		b.push_back(sz);
	} else {
		build(n - 1);
		++sz;
		a.push_back(sz);
		b.push_front(sz);
	}
}

int main() {
	scanf("%lld", &n);
	build(n);
	printf("%d\n", 2 * (int) a.size());

	for (int i : a) printf("%d ", i);
	for (int i : b) printf("%d ", i);
	puts("");
	return 0;
}
```