Problem:
Score : 1000 points

Problem Statement

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.

Sample Input 1

7

Sample Output 1

4
1 1 1 1

There are two good strings that appear as subsequences of s: (1,1) and (1,1,1,1). There are six occurrences of (1,1) and one occurrence of (1,1,1,1), for a total of seven.

Sample Input 2

299

Sample Output 2

23
32 11 11 73 45 8 11 83 83 8 45 32 32 10 100 73 32 83 45 73 32 11 10

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

//reference:wtl666wtl
# include <stdio.h>
int a[103], b[103];
int main()
{
    int p1=0, p2=0, m=100;
    long long n;
    scanf("%lld",&n);
    ++n;
    while(n > 1)
    {
        if(n&1) a[++p1] = m--, --n;
        else b[++p2] = m--, n>>=1;
    }
    printf("%d\n",p1+p2<<1);
    for(int i=1; i<=p1; ++i)
        printf("%d ",a[i]);
    for(int i=p2; i>=1; --i)
        printf("%d ",b[i]);
    for(int i=101-p1-p2; i<=100; ++i)
        printf("%d ",i);
    puts("");
    return 0;
}