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())
l=[]
e=""
r=[]
ans=[]

for _ in range(n):
    s=input()
    l.append(s)
if(n==1):
    for i in s:
        if(i=='?'):
            ans.append('c')
        else:
            ans.append(i)
else:
    for i in range(0,len(l[0])):
        for j in range(0,len(l)):
            e+=l[j][i]
        r.append(e)
        e=""
    for i in r:
        t=set(i)
        if ('?' in t ) and (len(t))>=3:
            ans.append('?')
        elif ('?' not in t) and (len(t))>=2:
            ans.append('?')
        elif(len(t)==1 and '?' not in t):
            ans.append(i[0])
        elif(len(t)==2 and '?' in t):
            for j in t:
                if(j=='?'):
                    continue
                else:
                    ans.append(j)
        elif(len(t)==1 and '?' in t):
            ans.append('x')

ans="".join(ans)
print(ans)
