#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define pf push_front
#define pb push_back
deque<int> A, B;
LL n;
int now = 1;

void work(LL x) {
    if (x == 1) return; else work(x / 2);
    A.pf(++now), B.pf(now);
    if (x & 1) A.pf(++now), B.pb(now);
}
 
int main() {
    cin >> n;
    work(n + 1);
    printf("%d\n", A.size() + B.size());
    for(int i = 0; i < A.size(); i++) printf("%d ", A[i]);
    for(int i = 0; i < B.size(); i++) printf("%d ", B[i]);
}
### 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;
#define pf push_front
#define pb push_back
deque<int> A, B;
LL n;
int now = 1;

void work(LL x) {
    if (x == 1) return; else work(x / 2);
    A.pf(++now), B.pf(now);
    if (x & 1) A.pf(++now), B.pb(now);
}
 
int main() {
    cin >> n;
    work(n + 1);
    printf("%d\n", A.size() + B.size());
    for(int i = 0; i < A.size(); i++) printf("%d ", A[i]);
    for(int i = 0; i < B.size(); i++) printf("%d ", B[i]);
}
```