You'll be given a list of two strings, and each will contain exactly one colon (`":"`) in the middle (but not at beginning or end). The length of the strings, before and after the colon, are random.

Your job is to return a list of two strings (in the same order as the original list), but with the characters after each colon swapped.


## Examples
```
["abc:123", "cde:456"]  -->  ["abc:456", "cde:123"]
["a:12345", "777:xyz"]  -->  ["a:xyz", "777:12345"]
```

Difficulty: introductory

Starter code:

def tail_swap(strings):
	

Solutions:
def tail_swap(strings):
    head0, tail0 = strings[0].split(':')
    head1, tail1 = strings[1].split(':')
    return [head0 + ':' + tail1, head1 + ':' + tail0]

def tail_swap(arr):
    fmt = '{}:{}'.format
    (head, tail), (head_2, tail_2) = (a.split(':') for a in arr)
    return [fmt(head, tail_2), fmt(head_2, tail)]


def tail_swap(strings):
    (h1, t1), (h2, t2) = (s.split(':') for s in strings)
    return [f"{h1}:{t2}", f"{h2}:{t1}"]

def tail_swap(ss):
  a = [s.split(':') for s in ss]
  a[0][1], a[1][1] = a[1][1], a[0][1]
  return [':'.join(p) for p in a]

def tail_swap(lst):
    (a,b),(c,d) = (s.split(':') for s in lst)
    return list(map(':'.join, ((a,d),(c,b))))

def tail_swap(s):
    h1,t1 = s[0].split(':')
    h2,t2 = s[1].split(':')
    return ['{}:{}'.format(h1,t2), '{}:{}'.format(h2,t1)]

def tail_swap(strings):
    s = [x.split(':') for x in strings]
    return [s[0][0] + ':' + s[1][1], s[1][0] + ':' + s[0][1]]

def tail_swap(strings):
    return '{0}:{3} {2}:{1}'.format(*[j for i in strings for j in i.split(':')]).split()

def tail_swap(strings):
    a, b = strings
    a, b = a.split(':'), b.split(':')
    return [f'{a[0]}:{b[1]}', f'{b[0]}:{a[1]}']

def tail_swap(strings):
    a,b = [ele.split(':') for ele in strings]
    return [ f'{a[0]}:{b[1]}',f'{b[0]}:{a[1]}' ]