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.
inp=input();loop=inp.count('#')
inp2=inp
crm,clm=0,0
cc=0
counter=inp.count(')')
crm+=inp2.count('(')
clm+=inp2.count(')')
l=[]
l2=[]
for i in range ( loop ):
    if inp2[0] == ')' or inp2[-1] == '(' or inp =='#' or inp.count('(')==0 or inp =='(#)' :
       print(-1)
       break
    elif clm+1<crm:
       inp2[inp2.find('#'):inp2.find('#')+1].replace('#','))')
       l.append(2)
       clm+=2
       l2.append('))')
    elif clm<crm:
       inp2[inp2.find('#'):inp2.find('#')+1].replace('#',')')
       l.append(1)
       clm+=1
       l2.append(')')
a=len(''.join(l2))
# print(counter + a )
# print(a)
# print(counter)
# print(crm)

for i in l:
   if crm != counter + a :
      print(-1)
   else :
      print ( i )
    