Python Itertools Code Optimisation

Given question (the contest is now over)

• a password consists of exactly n lowercase English letters.
• the password is melodious, meaning that consonants can only be next to vowels and vowels can only be next to consonants. Example: bawahaha
• the password cannot contain the letter y (because it's both a consonant and vowel).
• the first letter of the password can be either a vowel or consonant.

Given the length, n, of the password,

print all of the possible passwords that meet the conditions above.

Input Format

The line of input contains the integer (the length of the password).

Constraints

Output Format

Print each of the possible passwords, one per line. The order of the passwords does not matter.

My Code in Python:

import sys
import itertools

n = int(raw_input().strip())

consonants = ['b', 'c', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'm', 'n', 'p', 'q', 'r', 's', 't', 'v', 'w', 'x', 'z']
vowels = ['a', 'e', 'i', 'o', 'u']


test4 = ['b', 'c', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'm', 'n', 'p', 'q', 'r', 's', 't', 'v', 'w', 'x', 'z', 'a', 'e', 'i', 'o', 'u']
answer = set(itertools.product(test4, repeat=n))

for letters in answer:
    for j in xrange(len(letters)):
        flag = 1
        if j != len(letters) - 1:
            if letters[j] in vowels and letters[j+1] in vowels:
                flag = 0
                break
            if letters[j] in consonants and letters[j+1] in consonants:
                flag = 0
                break

    if flag:
        for j in letters:
            sys.stdout.write(j)
        print ""

Is there a better way to do this?