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())
output = ''
patterns = []
for _ in range(n):
    patterns.append(input())
for i in range(len(patterns[0])):
    chars, metaChars = '', ''
    for j in range(len(patterns)):
        if patterns[j][i] == '?':
            metaChars += '?'
        else:
            chars += patterns[j][i]
    if len(metaChars) == len(patterns):
        output += 'y'
    elif len(set(chars)) == 1:
        output += chars[0]
    else:
        output += '?'

print(output)
