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())
patterns = []
for _ in range(n):
    patterns.append(input())

# Find the pattern with minimum ?
min_qm = len(patterns[0])
for pat in patterns:
    qm = pat.count('?')
    min_qm = min(min_qm, qm)

# Find all patterns with this minimum ?
answer = []
for pat in patterns:
    qm = pat.count('?')
    if qm == min_qm:
        answer.append(pat)

# Find the intersection of all these patterns
intersection = []
for i in range(len(patterns[0])):
    c = answer[0][i]
    for a in answer[1:]:
        if a[i] != c:
            c = '?'
    intersection.append(c)

# Print the answer
print(''.join(intersection))