#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
ll n;

int main() {
    scanf("%lld", &n), n++;
    deque<ll> V;
    function<void(ll)> solve = [&](ll n) {
        if (n == 1) return;
        if (n & 1) solve(n - 1), V.push_front(V.size() + 1);
        else solve(n >> 1), V.push_back(V.size() + 1);
    };
    solve(n);
    for (int i = 1, j = V.size(); i <= j; i++) V.push_back(i);
    printf("%d\n", V.size());
    for (ll x : V) printf("%lld ", x);
    return 0;
}
### 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 <bits/stdc++.h>
using namespace std;

typedef long long ll;
ll n;

int main() {
    scanf("%lld", &n), n++;
    deque<ll> V;
    function<void(ll)> solve = [&](ll n) {
        if (n == 1) return;
        if (n & 1) solve(n - 1), V.push_front(V.size() + 1);
        else solve(n >> 1), V.push_back(V.size() + 1);
    };
    solve(n);
    for (int i = 1, j = V.size(); i <= j; i++) V.push_back(i);
    printf("%d\n", V.size());
    for (ll x : V) printf("%lld ", x);
    return 0;
}
```