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.
import sys
#sys.stdin=open('input.txt','r')
#sys.stdout=open('output.txt','w')

q=int(input())
r=[]

for i in range(q):
    s=input()
    r.append(list(s))

w=[[0 for i in range(q)] for j in range(len(r[0]))]
#print(w)
#print(r)
for i in range(0,len(r[0])):
    for j in range(0,q):
        #print(i,j,r[j][i])
        w[i][j]=r[j][i]
ans=[]
#print(w)
for i in range(len(w)):
    t=set(w[i])
    #print(t)
    n=len(t)
    if(n==1):
        if('?' in w[i]):
            ans.append('a')
        else:
            ans.append(w[i][0])
    if(n==2):
        if('?' in w[i]):
            for i in t:
                if(i!='?'):
                    ans.append(i)
        else:
            ans.append('?')
    elif(n>2):
        ans.append('?')

for  i in ans:
    print(i,end="")
