Dynamically build queryset - django

Question

I have these classes:
class Keyword(models.Model):
keyword = models.CharField(max_length=100, unique=True)
class Snippet(models.Model):
keywords = models.ManyToManyField(Keyword)
and a list of searched keywords:
searchlist = ['Photo','Cat']
I would like to dynamically query the database for snippets that contain both (AND operation) of the keywords in searchlist.
From that list of snippets, return the list of unique keywords from each of the snippets, excluding the original search terms.
So far I have this:
# Turn list of values into list of Q objects
queries = [Q(keywords__keyword__exact=tag) for tag in searchlist]
# Take one Q object from the list
query = queries.pop()
# Or the Q object with the ones remaining in the list
for item in queries:
query &= item
# Query the model
snippet_list = Snippet.objects.filter(query)
which returns the desired list of snippets but I'm unsure of how to retrieve the combined list of keywords, or if there is a more efficient way to achieve the end result.
EDIT:
I would like to achieve the following pseudo code on snippet_list:
for snippet in snippet_list:
combined_keywords += snippet.keywords
ordered_keyword_list = getOrderedKeywords() # <- I have this function already
final_list = intersection(ordered_keyword_list, combined_keywords)
where combined_keywords is all the keywords in the snippet_list. And final_list is a ordered unique version of combined_keywords.

Answer 1

The final solution looks like this thanks to #Soravux and #Dave-Webb. I'm not sure if this is the most efficient method, but it works.
# Turn list of values into list of Q objects
queries = [Q(keywords__keyword__exact=tag) for tag in valueslist]
query = Q()
for item in queries:
query &= item
# Query the model
snippet_list = Snippet.objects.filter(query)
combined_keywords = []
for snippet in snippet_list:
combined_keywords += snippet.keywords.all()
ordered_keyword_list = Keyword.objects.all().annotate(count=Count('snippet')).order_by('-count')
final_list = [x for x in ordered_keyword_list if x in combined_keywords]