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.
patterns_count = int(input())

patterns = []
for index in range(patterns_count):
	patterns.append(input())

best_intersect_pattern = []
for index in range(len(patterns[0])):
	letter = None
	for pattern in patterns:
		if pattern[index] != '?':
			if letter is None:
				letter = pattern[index]
			elif letter is not None and letter != pattern[index]:
				letter = '?'
				break
	if letter is None:
		letter = 'x'
	best_intersect_pattern.append(letter)

print(''.join(best_intersect_pattern))
 		 	 	 	  	    				  			 	 	