Problem

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  words, which might repeat, in word group . 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.

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Example

Group A contains 'a', 'b', 'a' Group B contains 'a', 'c'

For the first word in group B, 'a', it appears at positions 1 and 3 in group A. The second word, 'c', does not appear in group A, so print -1.

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Expected output:

1 3
-1

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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.

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Output Format

Output m lines.

The ith line should contain the 1-indexed positions of the occurrences of the ith word separated by spaces.

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Sample Input

STDIN   Function
-----   --------
5 2     group A size n = 5, group B size m = 2
a       group A contains 'a', 'a', 'b', 'a', 'b'
a
b
a
b
a       group B contains 'a', 'b'
b

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Sample Output

1 2 4
3 5

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Explanation

'a' appeared 3 times in positions 1, 2 and 4.

'b' appeared 2 times in positions 3 and 5.

In the sample problem, if 'c' also appeared in word group B, you would print -1.

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Solution

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from collections import defaultdict

n, m = list(map(int,input().split()))

d = defaultdict(list)

for i in range(n):
    d[input()].append(i+1)
    
for j in range(m):
    key = input()
    if len(d[key]) > 0:
        print(" ".join(str(k) for k in d[key]))
    else:
        print(-1)