extract should return all best scores

[mdziezyc]

Current:

>>> import rapidfuzz
>>> rapidfuzz.process.extract("aa", ["ab", "ba"], scorer=rapidfuzz.distance.Levenshtein.distance) 
[('ab', 1, 0), ('ba', 1, 1)]
>>> rapidfuzz.process.extract("aa", ["ab", "ba", "cc"], scorer=rapidfuzz.distance.Levenshtein.distance) 
[('ab', 1, 0), ('ba', 1, 1), ('cc', 2, 2)]
>>> rapidfuzz.process.extractOne("aa", ["ab", "ba", "cc"], scorer=rapidfuzz.distance.Levenshtein.distance) 
('ab', 1, 0)
>>> rapidfuzz.process.extract("aa", ["ab", "ba", "cc"], scorer=rapidfuzz.distance.Levenshtein.distance, limit=1) 
[('ab', 1, 0)]

Hypothetical/suggested:

>>> rapidfuzz.process.extract("aa", ["ab", "ba", "cc"], scorer=rapidfuzz.distance.Levenshtein.distance, limit="best") 
[('ab', 1, 0), ('ba', 1, 1)] # returns all with the lowest score

Together with #188 seems to me that it would be a killer option 🥇 to find all closest matches for the bulk queries lists.

Thanks for the great work!