#include <bits/stdc++.h>

using namespace std;
#define ll long long
const int N = 206;
int ans[N];

int main() {
    ll C; cin >> C;
    int HiBit = 63 - __builtin_clzll(++C), n = 0;
    for (int i = HiBit - 1; i >= 0; i--) {
        if (((C>>i)&1) == 0) n++, ans[n] = n;
        else {
            n++, ans[n] = n;
            for (int i = 1; i <= n; i++) ans[i]++;
            n++, ans[n] = 1;
        }
    }
    cout << 2 * n << "\n";
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
    for (int i = 1; i <= n; i++) cout << i << " ";
    cout << "\n";

    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
#define ll long long
const int N = 206;
int ans[N];

int main() {
    ll C; cin >> C;
    int HiBit = 63 - __builtin_clzll(++C), n = 0;
    for (int i = HiBit - 1; i >= 0; i--) {
        if (((C>>i)&1) == 0) n++, ans[n] = n;
        else {
            n++, ans[n] = n;
            for (int i = 1; i <= n; i++) ans[i]++;
            n++, ans[n] = 1;
        }
    }
    cout << 2 * n << "\n";
    for (int i = 1; i <= n; i++) cout << ans[i] << " ";
    for (int i = 1; i <= n; i++) cout << i << " ";
    cout << "\n";

    return 0;
}

```