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.
l = []
for _ in range(int(input())): l.append(input())
l = [e for e in zip(*l)]
s = ""
mark = False
letter = ''
for i in range(len(l)):
    if all(list(map(lambda x: x == l[i][0] and x != '?', l[i]))):
        s += l[i][0]
        continue
    for j in range(len(l[0])):
        if l[i][j] != '?':
            if not mark or l[i][j] == letter:
                mark = True
                letter = l[i][j]
                # print(letter)
            else:
                s+='?'
                break
    else:
        s+=letter if letter != '' else 'a'
print(s)