Write a function that takes a list comprised of other lists of integers and returns the sum of all numbers that appear in two or more lists in the input list.  Now that might have sounded confusing, it isn't:
```python
repeat_sum([[1, 2, 3],[2, 8, 9],[7, 123, 8]])
>>> sum of [2, 8]
return 10

repeat_sum([[1], [2], [3, 4, 4, 4], [123456789]])
>>> sum of []
return 0

repeat_sum([[1, 8, 8], [8, 8, 8], [8, 8, 8, 1]])
sum of [1,8]
return 9
```

Source: codewars | Difficulty: EASY
Tags: Data structures

Starter code: def repeat_sum(l):
	

Solutions:
from collections import defaultdict

def repeat_sum(l):
	count = defaultdict(int)
	for l1 in l:
		for val in set(l1):
			count[val] += 1
	return sum((k for (k, v) in list(count.items()) if v > 1))


from collections import Counter

def repeat_sum(l):
	counter = sum((Counter(set(el)) for el in l), Counter())
	return sum((num for num in counter if counter[num] > 1))


def repeat_sum(l):
	(s, r) = (set(), set())
	for a in map(set, l):
		r |= a & s
		s |= a
	return sum(r)


from collections import Counter
from itertools import chain

def repeat_sum(l):
	return sum((key for (key, value) in Counter(chain.from_iterable((set(x) for x in l))).items() if value > 1))


from collections import Counter as Cnt
repeat_sum = lambda l: sum((k for (k, v) in Cnt([e for ll in l for e in set(ll)]).items() if v > 1))


def repeat_sum(l):
	f = [n for s in l for n in set(s)]
	return sum((n for n in set(f) if f.count(n) > 1))


from collections import Counter

def repeat_sum(l):
	commons = Counter()
	for s in l:
		commons.update(set(s))
	return sum((x for (x, c) in commons.items() if c > 1))


repeat_sum = lambda x: sum(list(set([j for i in x for j in i if sum((1 for q in x if j in q)) > 1])))