#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int main(void) {
    ll N, tmp;
    int num = 0;
    cin >> N;
    tmp = N + 1;
    vector<int> ans;

    while(tmp) {tmp /= 2; num++;}
    for(int i=0; i<num-1; i++) {
        if(((N+1) >> i) & 1) ans.emplace_back(100-i);
        ans.emplace_back(i+1);
    }

    printf("%d\n", ans.size()+100);
    for(int a : ans) printf("%d ", a);
    for(int i=1; i<=100; i++) printf("%d ", i);
    printf("\n");
    return 0;
}
### Prompt

Generate a CPP solution 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;
typedef long long ll;

int main(void) {
    ll N, tmp;
    int num = 0;
    cin >> N;
    tmp = N + 1;
    vector<int> ans;

    while(tmp) {tmp /= 2; num++;}
    for(int i=0; i<num-1; i++) {
        if(((N+1) >> i) & 1) ans.emplace_back(100-i);
        ans.emplace_back(i+1);
    }

    printf("%d\n", ans.size()+100);
    for(int a : ans) printf("%d ", a);
    for(int i=1; i<=100; i++) printf("%d ", i);
    printf("\n");
    return 0;
}
```