#include <cstdio>
#include <vector>
#define ll long long
#define rep(i, j, k) for (i=j;i<=k;i++)
using namespace std;
ll n;
int i, k;
vector<int> a;

void solve(ll n, int &k) {
    if (n == 1) {
        a.push_back(1);
        k = 1;
        return;
    }
    if (n & 1) {
        solve((n - 1) / 2, k);
        k++;
        a.push_back(k);
    } else {
        solve(n - 1, k);
        k++;
        a.insert(a.begin(), k);
    }
}

int main() {
    scanf("%lld", &n);
    solve(n, k);
    rep(i, 1, k) a.push_back(i);
    printf("%d\n", a.size());
    rep(i, 0, a.size() - 1) printf("%d ", a[i]);
    return 0;
}
### Prompt

Your challenge is to write 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 <cstdio>
#include <vector>
#define ll long long
#define rep(i, j, k) for (i=j;i<=k;i++)
using namespace std;
ll n;
int i, k;
vector<int> a;

void solve(ll n, int &k) {
    if (n == 1) {
        a.push_back(1);
        k = 1;
        return;
    }
    if (n & 1) {
        solve((n - 1) / 2, k);
        k++;
        a.push_back(k);
    } else {
        solve(n - 1, k);
        k++;
        a.insert(a.begin(), k);
    }
}

int main() {
    scanf("%lld", &n);
    solve(n, k);
    rep(i, 1, k) a.push_back(i);
    printf("%d\n", a.size());
    rep(i, 0, a.size() - 1) printf("%d ", a[i]);
    return 0;
}
```