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())
s = input()
str_len = len(s)
good = []
for ch in s:
    if ch == '?':
        good.append('$')
    else:
        good.append(ch)
r = range(n - 1)
for i in r:
    s = input()
    j = int(0)
    while j < str_len:
        if (s[j] == '?') and (good[j] == '$'):
            good[j] = '$';
        elif s[j] == '?':
            good[j] = good[j]
        elif good[j] == '$':
            good[j] = s[j]
        elif not(s[j] == good[j]):
            good[j] = '?'
        j = j + 1

ans = ""
for ch in good:
    if ch == '$':
        ans = ans + 'z'
    else:
        ans = ans + ch
    
print(ans)
