def unique(s):
    #taken from http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/52560
    """Return a list of the elements in s, but without duplicates"""

    n = len(s)
    if n == 0:
        return []

    # Try using a dict first, as that's the fastest and will usually
    # work.  If it doesn't work, it will usually fail quickly, so it
    # usually doesn't cost much to *try* it.  It requires that all the
    # sequence elements be hashable, and support equality comparison.
    u = {}
    try:
        for x in s:
            u[x] = 1
    except TypeError:
        del u  # move on to the next method
    else:
        return u.keys()

    # We can't hash all the elements.  Second fastest is to sort,
    # which brings the equal elements together; then duplicates are
    # easy to weed out in a single pass.
    # NOTE:  Python's list.sort() was designed to be efficient in the
    # presence of many duplicate elements.  This isn't true of all
    # sort functions in all languages or libraries, so this approach
    # is more effective in Python than it may be elsewhere.
    try:
        t = list(s)
        t.sort()
    except TypeError:
        del t  # move on to the next method
    else:
        assert n > 0
        last = t[0]
        lasti = i = 1
        while i < n:
            if t[i] != last:
                t[lasti] = last = t[i]
                lasti += 1
            i += 1
        return t[:lasti]

    # Brute force is all that's left.
    u = []
    for x in s:
        if x not in u:
            u.append(x)
    return u

def append_if_new(x, l):
        if not x in l:
                l.append(x)