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 Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program tautonym_puzzle
  implicit none
  integer(8) :: n
  integer :: k = 40, p(80) = 0, i
  read(*,*) n
  do while (k > 0)
    if (n >= lshift(1_8,k)-1_8) exit
    k = k-1
  end do
  do i = 1, k
    p(i) = i
  end do
  n = n-lshift(1_8,k)+1_8
  do i = k-1, 0, -1
    if (n < lshift(1_8,i)) cycle
    p(i+1:k+1) = p(i:k)
    k = k+1
    p(i) = k
    n = n-lshift(1_8,i)
  end do
  do i = 1, k
    p(k+i) = i
  end do
  write(*,'(i0)') 2*k
  write(*,'(i0)',advance='no') p(1)
  do i = 2, 2*k
    write(*,'(x,i0)',advance='no') p(i)
  end do
  write(*,*)
end program tautonym_puzzle