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.

n = int(input())
patterns = []
for i in range (n):
  patterns.append(input())

size = len(patterns[0])
result = [0] * size

for i in range (len(patterns[0])):
  temp = []
  
  for padrao in patterns:
    temp.append(padrao[i])
  
  if temp.count('?') == n:
    result[i] = 'x'
  else:
    temp = list(filter(lambda x: x != '?', temp))
    if(temp.count(temp[0]) == len(temp)):
      result[i] = temp[0]
    else:
      result[i] = '?'

print(*result, sep='')
 		 		 		 	  			  			 	 						