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)

# include <stdio.h>
int x[101], y[101];
int main()
{    
int a=0, b=0, m=100,i;    
long long n;    
scanf("%lld",&n);    
++n;    
while(n>1)    
{        
if(n%2) 
{
a++;
x[a]=m;
m--;
n--;    
}
else 
{
b++;
y[b]=m;
m--;
n/=2;    
}   
}
printf("%d\n",(a+b)*2);    
for(i=1;i<=a;i++)        
printf("%d ",x[i]);    
for(i=b;i>=1;i--)        
printf("%d ",y[i]);    
for(i=101-a-b;i<=100;i++)        
printf("%d ",i);   
return 0;
}