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.
from collections import Counter

def match(str1,str2):
    answ = ""
    if str2:
        for i in range(len(str2)):
            if str2[i] != str1[i] and str1[i] == '?':
                answ += str2[i]
            elif str2[i] != str1[i] and str2[i] == '?':
                answ += str1[i]
            elif str1[i] != str2[i]:
                answ += "?"
            elif str1[i] == str2[i] and str1[i] != '?':
                answ += str2[i]
            else:
                answ += "a"
    else:
        for char in str1:
            if char == '?':
                answ += 'a'
            else:
                answ += char
    return answ

def main():
    n = int(input())
    if n >= 2:
        s1 = input()
        s2 = input()
        for i in range(2,n):
            res = input()
        res = match(s1,s2)
        print(res)

    else:
        s1 = input()
        res = match(s1,"")
        print(res)

main()
