#include <bits/stdc++.h> 
using namespace std; 
long long N; 
int p[200], q[200], cnt = 1;
void solve(long long n) {
	if (!n) return;
   	if (~n & 1) 
		q[++q[0]] = cnt++, solve(n - 1); 
   	else
   		p[++p[0]] = cnt++, solve(n >> 1); 
} 
int main () { 
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin); 
#endif 
	cin >> N; 
	solve(N); 
	printf("%d\n", cnt + 100 - 1);
	for (int i = 1; i <= q[0]; i++) printf("%d ", cnt - q[i]); 
	for (int i = p[0]; i; i--) printf("%d ", cnt - p[i]); 
	for (int i = 1; i <= 100; i++) printf("%d ", i); 
	putchar('\n'); 
    return 0; 
} 
### Prompt

Develop a solution in Cpp 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; 
long long N; 
int p[200], q[200], cnt = 1;
void solve(long long n) {
	if (!n) return;
   	if (~n & 1) 
		q[++q[0]] = cnt++, solve(n - 1); 
   	else
   		p[++p[0]] = cnt++, solve(n >> 1); 
} 
int main () { 
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin); 
#endif 
	cin >> N; 
	solve(N); 
	printf("%d\n", cnt + 100 - 1);
	for (int i = 1; i <= q[0]; i++) printf("%d ", cnt - q[i]); 
	for (int i = p[0]; i; i--) printf("%d ", cnt - p[i]); 
	for (int i = 1; i <= 100; i++) printf("%d ", i); 
	putchar('\n'); 
    return 0; 
} 
```