#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n, u, d;
vector<ll> x, ans;

int main()
{
	scanf("%lld",&n);
	while(n > 1) {
		if(n%2) {x.push_back(1); n /= 2;}
		else {x.push_back(0); n--;}
	}
	reverse(x.begin(), x.end());
	ans.push_back(0);
	for(auto &T : x) {
		if(T) ans.push_back(++u);
		else ans.push_back(--d);
	}
	printf("%d\n",2*ans.size());
	for(ll i=0;i<ans.size();i++) printf("%lld ", i+1);
	for(ll i=0;i<ans.size();i++) printf("%lld ", ans[i]-d+1);
}

### Prompt

Create a solution in Cpp for 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;
typedef long long ll;
ll n, u, d;
vector<ll> x, ans;

int main()
{
	scanf("%lld",&n);
	while(n > 1) {
		if(n%2) {x.push_back(1); n /= 2;}
		else {x.push_back(0); n--;}
	}
	reverse(x.begin(), x.end());
	ans.push_back(0);
	for(auto &T : x) {
		if(T) ans.push_back(++u);
		else ans.push_back(--d);
	}
	printf("%d\n",2*ans.size());
	for(ll i=0;i<ans.size();i++) printf("%lld ", i+1);
	for(ll i=0;i<ans.size();i++) printf("%lld ", ans[i]-d+1);
}

```