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.
a=input()
l=list(a)
q=[]
t=0# )数目
b=0# (数目
c=0# #数目
g=0
for i in range(len(l)):
        if l[i]==')':
            t+=1
            if t>i:
                g=-1
                break
           
        elif l[i]=='(':
             b+=1
        elif l[i]=='#':
             c+=1
if t>=b:
    g=-1
if g==-1:
    print(g)
else:

 p=[0]*(len(l))
 if l[0]=='(':
  p[0]=1
 elif l[0]=='#':
    p[0]=0
 for i in range(1,len(l)):
    if l[i]=='(':
        p[i]+=(p[i-1]+1)
    elif l[i]==')':
        p[i]+=(p[i-1]-1)
    elif l[i]=='#':
        p[i]+=p[i-1]
 for i in range(len(l)):
    if l[i]=='#':
        q.append(p[i])
 sum=b-t
 o=0
 v=[0]*(len(q))
 for i in range(0,len(q)-1):
    v[i]=q[i+1]-q[i]
    o+=v[i]
 v[-1]=sum-o
 for i in range(len(v)):
    print(v[i])
    
        
                
