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 = []

if n == 1:
    pattern = input()
    
else:
    for _ in range(n):
        pattern = input()
        # if we have more than one pattern
        if output:
            for i in range(len(pattern)):
                # check if pattern match ??? or aaa
                if output[i] == pattern[i]:
                    continue
                # check if pattern match *a or *?
                elif output[i] == '*':
                    if pattern[i] == '?':
                        continue
                    else:
                        output[i] = pattern[i]
                # check if pattern match ?
                elif output[i] == '?':
                     continue
                # check if the current pattern is different from the comulative pattern
                elif output[i] != pattern[i] != '?':
                    output[i] = '?'          
        else:
            for i in range(len(pattern)):
                # * is special character which replaced later for pattern optimization
                if pattern[i] == '?':
                    output.append('*')
                else:
                    output.append(pattern[i])
        

print(''.join(map(lambda char: 'y' if char == '*' else char, output)))
