#include<cstdio>
long long n;
int w[100], cnt, R[200], cc, RC;
int main(){
    scanf("%lld",&n);
    n++;
    while(n){
        w[cnt++] = n%2;
        n/=2;
    }
    int p1 = 50, p2 = 51, p3 = 101, i;
    for(i=cnt-2;i>=0;i--){
        cc++;
        R[p2] = cc;
        R[p3] = R[p2];
        if(w[i]){
            R[++p3] = ++cc;
            R[p1--] = cc;
        }
        p2++,p3++;
    }
    for(i=1;i<200;i++)if(R[i])RC++;
    printf("%d\n",RC);
    for(i=1;i<200;i++)if(R[i])printf("%d ",R[i]);
}
### Prompt

Please formulate 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>
long long n;
int w[100], cnt, R[200], cc, RC;
int main(){
    scanf("%lld",&n);
    n++;
    while(n){
        w[cnt++] = n%2;
        n/=2;
    }
    int p1 = 50, p2 = 51, p3 = 101, i;
    for(i=cnt-2;i>=0;i--){
        cc++;
        R[p2] = cc;
        R[p3] = R[p2];
        if(w[i]){
            R[++p3] = ++cc;
            R[p1--] = cc;
        }
        p2++,p3++;
    }
    for(i=1;i<200;i++)if(R[i])RC++;
    printf("%d\n",RC);
    for(i=1;i<200;i++)if(R[i])printf("%d ",R[i]);
}
```