Malek has recently found a treasure map. While he was looking for a treasure he found a locked door. There was a string *s* written on the door consisting of characters '(', ')' and '#'. Below there was a manual on how to open the door. After spending a long time Malek managed to decode the manual and found out that the goal is to replace each '#' with one or more ')' characters so that the final string becomes beautiful. 

Below there was also written that a string is called beautiful if for each *i* (1 ≤ *i* ≤ |*s*|) there are no more ')' characters than '(' characters among the first *i* characters of *s* and also the total number of '(' characters is equal to the total number of ')' characters. 

Help Malek open the door by telling him for each '#' character how many ')' characters he must replace it with.
The first line of the input contains a string *s* (1 ≤ |*s*| ≤ 10<sup class="upper-index">5</sup>). Each character of this string is one of the characters '(', ')' or '#'. It is guaranteed that *s* contains at least one '#' character.

from sys import stdin

s = stdin.readline().strip()

def solve(s):
	idx = s.index('#')
	totalNum = s.count('#')
	left, right = s[: idx].count('('), s[: idx].count(')')
	while totalNum > 0:
		if right >= left:
			print(-1)
			return
		if totalNum > 1:
			replace = 1
			prev, idx = idx, s.index('#', idx + 1)
			left += s[prev : idx].count('(')
			right += (s[prev : idx].count(')') + replace)
		else:
			left += s[idx :].count('(')
			right += s[idx :].count(')')
			replace = left - right if left > right else -1
		print(replace)
		totalNum -= 1

solve(s)

    		  		     		 			  	     		