#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL n;
int num[1111];
int main(){
    cin>>n;
    n++;
    int hd, tl;
    hd = 500;
    tl = hd - 1;
    int ma = 0;
    for(int i = 0;i <= 40; i++)if(n >> i & 1)ma = i;
    int tot = 1;
    for(int i = ma - 1;i >= 0; i--){
        num[++tl] = tot;
        tot++;
        if(n >> i & 1)num[--hd] = tot, tot++;
    }
    printf("%d\n", tot * 2 - 1);
    for(int i = hd;i <= tl; i++)printf("%d ", num[i]);
    for(int i = 1;i <= tot; i++)printf("%d ", i);
    return 0;
}

### Prompt

Your task is to create 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;
#define LL long long
LL n;
int num[1111];
int main(){
    cin>>n;
    n++;
    int hd, tl;
    hd = 500;
    tl = hd - 1;
    int ma = 0;
    for(int i = 0;i <= 40; i++)if(n >> i & 1)ma = i;
    int tot = 1;
    for(int i = ma - 1;i >= 0; i--){
        num[++tl] = tot;
        tot++;
        if(n >> i & 1)num[--hd] = tot, tot++;
    }
    printf("%d\n", tot * 2 - 1);
    for(int i = hd;i <= tl; i++)printf("%d ", num[i]);
    for(int i = 1;i <= tot; i++)printf("%d ", i);
    return 0;
}

```