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

deque<int> ans;
int m;

void DFS(long long n) {
    if (n == 1) return ;
    if (n & 1) {
        DFS(n-1);
        ans.push_front(++m);
    } else {
        DFS(n>>1);
        ans.push_back(++m);
    }
}

int main() {
    long long n;
    scanf("%lld", &n);
    m = 0;
    DFS(n+1);
    printf("%d\n", m*2);
    for (int x: ans) {
        printf("%d ", x);
    }
    for (int i=1; i<=m; ++i) {
        printf("%d%c", i, i==m ? '\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;

deque<int> ans;
int m;

void DFS(long long n) {
    if (n == 1) return ;
    if (n & 1) {
        DFS(n-1);
        ans.push_front(++m);
    } else {
        DFS(n>>1);
        ans.push_back(++m);
    }
}

int main() {
    long long n;
    scanf("%lld", &n);
    m = 0;
    DFS(n+1);
    printf("%d\n", m*2);
    for (int x: ans) {
        printf("%d ", x);
    }
    for (int i=1; i<=m; ++i) {
        printf("%d%c", i, i==m ? '\n' : ' ');
    }
    return 0;
}

```