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
input = sys.stdin.readline

############ ---- Input Functions ---- ############
inp = sys.stdin.readline
def input(): return inp().strip()
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))


def solve():
    n = ii()
    patter = input()
    for _ in range(n-1):
        curr = list(input())
        index = 0
        for p, c in zip(patter, curr):
            # print(p, c)
            if p != c:
                curr[index] = '*'
            if p == "?" and c != "*":
                curr[index] = c
            if p != "*" and c == "?":
                curr[index] = p
            index += 1
        patter = curr[:]
    # print(patter)
    for i in patter:
        if i == "*":
            print('?', end="")
        elif i == "?":
            print('a', end="")
        else:
            print(i, end="")


def main():
    solve()


if __name__ == '__main__':
    main()
