#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(V) V.begin(),V.end()
#define MOD 1000000007

int main() {
	ll N;
	cin >> N;
	ll d = 1;
	while (1) {
		if (N / (1LL << d) < 1LL) {
			break;
		}
		d++;
	}
	d--;
	N -= (1LL << d);
	ll ad = 99;
	queue<int> Q;
	Q.push(100);
	for (ll i = 0;;i++) {
		if (N & (1LL << i)) {
			Q.push(ad);
			ad--;
		}
		if (i == d)break;
		Q.push(i + 1);
	}
	for (int i = 1;i <= 100;i++) {
		Q.push(i);
	}
	cout << Q.size() << endl;
	while (!Q.empty()) {
		cout << Q.front() << " ";
		Q.pop();
	}
	cout << endl;
}
### Prompt

Please create a solution in Cpp 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 <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(V) V.begin(),V.end()
#define MOD 1000000007

int main() {
	ll N;
	cin >> N;
	ll d = 1;
	while (1) {
		if (N / (1LL << d) < 1LL) {
			break;
		}
		d++;
	}
	d--;
	N -= (1LL << d);
	ll ad = 99;
	queue<int> Q;
	Q.push(100);
	for (ll i = 0;;i++) {
		if (N & (1LL << i)) {
			Q.push(ad);
			ad--;
		}
		if (i == d)break;
		Q.push(i + 1);
	}
	for (int i = 1;i <= 100;i++) {
		Q.push(i);
	}
	cout << Q.size() << endl;
	while (!Q.empty()) {
		cout << Q.front() << " ";
		Q.pop();
	}
	cout << endl;
}
```