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# Uses python3
import sys


#this is a dynamic programming approach in an iterative function
def optimal_sequence(n):
    array_previous_numbers = [None]*(n+1)
    array_min_ops = [0]+[None]*n

    for number in range(1, n+1):
        previous_number = number - 1
        min_ops = array_min_ops[previous_number]+1
        # this operation records the value of computing optimal_sequence by adding 1 to the previous number.
        # this answer is used as the default fill in value until a better operation is found.
        # the next two if statesments

        if number % 3 == 0:
            candidate_previous_number = number // 3
            candidate_num_ops = array_min_ops[candidate_previous_number]+1
            if candidate_num_ops < min_ops:
                previous_number = candidate_previous_number
                min_ops = candidate_num_ops

        if number % 2 == 0:
            candidate_previous_number = number // 2
            candidate_num_ops = array_min_ops[candidate_previous_number] + 1
            if candidate_num_ops < min_ops:
                previous_number = candidate_previous_number
                min_ops = candidate_num_ops

        array_previous_numbers[number], array_min_ops[number] = previous_number, min_ops
        # this fills the two empty arrays with the previous numbers and min_ops

    # this is an ingenious section to reconstruct the solution
    numbers = []
    countdown = n
    while countdown > 0:
        numbers.append(countdown)
        countdown = array_previous_numbers[countdown]
    numbers.reverse()
    return numbers


#this is an recursive version which fails with very big numbers.
# array = {}
# def optimal_sequence(n):
#     if n == 1:
#         array[n] = 1, -1
# #        return 1, -1
#     # if 1 is reached, terminate
#     if array.get(n) is not None:
#         return array[n]
#     # if the indice in the array is not empty return the answer
#
#     ans = (optimal_sequence(n-1)[0]+1, n-1)
#     # this operation records the value of computing optimal_sequence by adding 1 to the previous number.
#     # this answer is used as the default fill in value until a better operation is found.
#     # returns a tuple, (a,b) where a is the number of operations and b is the previous number used to get the solution
#     # the below solutions records the value of computing optimal sequences by dividing by two or 3
#
#     if n % 2 == 0:
#         ret = optimal_sequence(n//2)
#         if ans[0]>ret[0]:
#             ans = (ret[0]+1, n//2)
#     if n % 3 == 0:
#         ret = optimal_sequence(n // 3)
#         if ans[0] > ret[0]:
#             ans = (ret[0]+1, n // 3)
#
#     array[n] = ans
#     return ans


#greedy solution
# def optimal_sequence(n):
#     sequence = []
#     while n >= 1:
#         sequence.append(n)
#         if n % 3 == 0:
#             n = n // 3
#         elif n % 2 == 0:
#             n = n // 2
#         else:
#             n = n - 1
#     return reversed(sequence)

input = sys.stdin.read()
n = int(input)
sequence = list(optimal_sequence(n))
print(len(sequence) - 1)
for x in sequence:
    print(x, end=' ')