Problem: Steve Jobs' ghost has come down to BITS Pilani Hyderabad Campus, since he has heard a lot about the coding culture here. Hoping to recruit talented people to work on top-secret Apple software, he keeps a strict coding test.

Apple is currently working on a upgraded version of a automated form filler. They call it the iComplete application. Steve Jobs is looking for BPHC talent to work on it.

Therefore, his coding test requires contestants to work on a basic iComplete application.

iComplete V1.0 is an application that works by receiving part of a word and automatically completing it depending on the dictionary present on the device. Therefore for the coding test, you are given a dictionary of n words. Now Steve Jobs will come over to your machine and type a string s. Your task is to complete s to match one of the words in the dictionary.

In case many words have s as a prefix, you have to find the lexicographically smallest word amongst these. In case no match is found, simply return the string s.

Input

The first line contains the string s which Steve Jobs' ghost types. The second line contains an integer n (1 ≤ n ≤ 100) which is the number of words in the dictionary. Then follow n lines which are the words on the dictionary, one on each line. All the lines have lengths of 1 to 100 characters inclusively, and consist of lowercase English letters only.
Output
If s is not the beginning of any of n words in the dictionary, print s. Otherwise, print the lexicographically lowest word beginning with s.

The lexicographical order is the order of words in a dictionary.

Sample test(s)
Input
next
2
nextpermutation
nextelement

Output
nextelement

Input
find
4
find
findfirstof
findit
fand

Output
find

Input
find
4
fondfind
fondfirstof
fondit
fand

Output
find

SAMPLE INPUT
next
2
nextpermutation
nextelement

SAMPLE OUTPUT
nextelement

Explanation

Since nextelement is smaller when seen in dictionary, it is preferred over nextpermutation.

Solution (in PYTHON):

Solution:
def main():
    string = raw_input()
    dict_count = input()
    user_dict = list()
    res = None 
    for each in xrange(dict_count):
        user_dict.append(raw_input())
    for each in sorted(user_dict):
        if each.startswith(string):
            res = each
            break
    print res if res else string

if __name__ == '__main__':
    main()