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:
w=raw_input()
n=input()
l=list()
for i in range(0,n):
	a=raw_input()
	l.append(a)

le=0
s=-1
for j in range(0,n):
	if w==str(l[j]):
		s=j
		break
	if(str(l[j]).find(w,0,len(l[j]))==0):
		if(le==0):
			
			le=len(str(l[j]))
			s=j
		else:
			if(le>len(l[j])):
				le=len(l[j]);
				s=j
		
if s==-1:
	print w
else:
	print l[s]