Developers often face with regular expression patterns. A pattern is usually defined as a string consisting of characters and metacharacters that sets the rules for your search. These patterns are most often used to check whether a particular string meets the certain rules.

In this task, a pattern will be a string consisting of small English letters and question marks ('?'). The question mark in the pattern is a metacharacter that denotes an arbitrary small letter of the English alphabet. We will assume that a string matches the pattern if we can transform the string into the pattern by replacing the question marks by the appropriate characters. For example, string aba matches patterns: ???, ??a, a?a, aba.

Programmers that work for the R1 company love puzzling each other (and themselves) with riddles. One of them is as follows: you are given *n* patterns of the same length, you need to find a pattern that contains as few question marks as possible, and intersects with each of the given patterns. Two patterns intersect if there is a string that matches both the first and the second pattern. Can you solve this riddle?
The first line contains a single integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>) — the number of patterns. Next *n* lines contain the patterns.

It is guaranteed that the patterns can only consist of small English letters and symbols '?'. All patterns are non-empty and have the same length. The total length of all the patterns does not exceed 10<sup class="upper-index">5</sup> characters.
testes = int(input())
paterns = []
resposta = ""
for i in range (testes):
    inp = str(input())
    paterns.append(inp)

for i in range (len(paterns[0])):
    resp = ""
    for p in paterns:
        if resp == "":
            if p[i] != "?":
                resp = p[i]
        else:
            if p[i] != resp and p[i] != "?":
                resp = "?"
                break
        
    if resp == "":
        resposta = resposta + "x"   
    else:
        resposta = resposta + resp

print(resposta)

			 	  		  		  		  	 	 	  			 	