#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
template<typename T> inline void chkmin(T &a, const T &b) { a = a < b ? a : b; }
template<typename T> inline void chkmax(T &a, const T &b) { a = a > b ? a : b; }

int arr[205];
LL n;

int main() {
	scanf("%lld", &n); ++n;
	int he = 100, ta = 99, tn = 0, c = 0;
	for (LL i = n; i; i >>= 1) ++tn;
	for (int i = tn - 2; i >= 0; i--) {
		arr[++ta] = ++c;
		if (n >> i & 1) arr[--he] = ++c;
	}
	printf("%d\n", c << 1);
	for (int i = he; i <= ta; i++) printf("%d ", arr[i]);
	for (int i = 1; i <= c; i++) printf("%d ", i);
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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;
typedef long long LL;
typedef pair<int, int> P;
template<typename T> inline void chkmin(T &a, const T &b) { a = a < b ? a : b; }
template<typename T> inline void chkmax(T &a, const T &b) { a = a > b ? a : b; }

int arr[205];
LL n;

int main() {
	scanf("%lld", &n); ++n;
	int he = 100, ta = 99, tn = 0, c = 0;
	for (LL i = n; i; i >>= 1) ++tn;
	for (int i = tn - 2; i >= 0; i--) {
		arr[++ta] = ++c;
		if (n >> i & 1) arr[--he] = ++c;
	}
	printf("%d\n", c << 1);
	for (int i = he; i <= ta; i++) printf("%d ", arr[i]);
	for (int i = 1; i <= c; i++) printf("%d ", i);
	return 0;
}
```