=====Problem Statement=====
The defaultdict tool is a container in the collections class of Python. It's similar to the usual dictionary (dict) container, but the only difference is that a defaultdict will have a default value if that key has not been set yet. If you didn't use a defaultdict you'd have to check to see if that key exists, and if it doesn't, set it to what you want.
For example:
from collections import defaultdict
d = defaultdict(list)
d['python'].append("awesome")
d['something-else'].append("not relevant")
d['python'].append("language")
for i in d.items():
    print i
This prints:
('python', ['awesome', 'language'])
('something-else', ['not relevant'])
In this challenge, you will be given 2 integers, n and m. There are n words, which might repeat, in word group A. There are m words belonging to word group B. For each m words, check whether the word has appeared in group A or not. Print the indices of each occurrence of m in group A. If it does not appear, print -1.
=====Constraints=====
1≤n≤10000
1≤m≤100
1≤length of each word in the input≤100
=====Input Format=====
The first line contains integers, n and m separated by a space.
The next n lines contains the words belonging to group A.
The next m lines contains the words belonging to group B.
=====Output Format=====
Output m lines.
The ith line should contain the 1-indexed positions of the occurrences of the ith word separated by spaces.
from collections import defaultdict
d = defaultdict(list)
n,m=list(map(int,input().split()))
for i in range(n):
    w = input()
    d[w].append(str(i+1))
for j in range(m):
    w = input()
    print((' '.join(d[w]) or -1))

from collections import defaultdict
d = defaultdict(list)
n,m=list(map(int,input().split()))
for i in range(n):
    w = input()
    d[w].append(str(i+1))
for j in range(m):
    w = input()
    print((' '.join(d[w]) or -1))
