#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;

LL a[1005],b[1005],cna = 0,cnb = 0,tot = 1;
LL n,c[1005],cnc = 0;

int main(){
	ios::sync_with_stdio(false);
	cin >> n; n ++;
	while(n){
		c[++ cnc] = n & 1;
		n >>= 1;
	}
	
	b[++ cnb] = 1;
	for(LL i = cnc - 1;i >= 1;i --){
		if(c[i]) a[++ cna] = ++ tot;
		if(i > 1) b[++ cnb] = ++ tot;
	}
	
	cout << tot + 100 << endl;
	for(LL i = cna;i >= 1;i --) cout << a[i] << ' ';
	for(LL i = 1;i <= cnb;i ++) cout << b[i] << ' ';
	for(LL i = 1;i <= 100;i ++) cout << i << (i == 100 ? '\n' : ' ');
	return 0;
}
// 太难了
### 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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;

LL a[1005],b[1005],cna = 0,cnb = 0,tot = 1;
LL n,c[1005],cnc = 0;

int main(){
	ios::sync_with_stdio(false);
	cin >> n; n ++;
	while(n){
		c[++ cnc] = n & 1;
		n >>= 1;
	}
	
	b[++ cnb] = 1;
	for(LL i = cnc - 1;i >= 1;i --){
		if(c[i]) a[++ cna] = ++ tot;
		if(i > 1) b[++ cnb] = ++ tot;
	}
	
	cout << tot + 100 << endl;
	for(LL i = cna;i >= 1;i --) cout << a[i] << ' ';
	for(LL i = 1;i <= cnb;i ++) cout << b[i] << ' ';
	for(LL i = 1;i <= 100;i ++) cout << i << (i == 100 ? '\n' : ' ');
	return 0;
}
// 太难了
```