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.
n=input()
a=[0]
p=n.count('#')
b=n.count('(')
c=n.count(')')
s=n[::-1].index('#')
s=len(n)-1-s
for j in range(s):
    if n[j]=='(':
        a.append(a[-1]+1)
    elif n[j]==')':
        a.append(a[-1]-1)
    elif n[j]=='#':
        a.append(a[-1]-1)
        n.replace('#',')')
n.replace('#',')'*(b-c-p+1))
if s==len(n)-1:
    if min(a)<0:
        print(-1)
    else:
        for _ in range(p-1):
            print(1)
        print(b-c-p+1)
else:
    for j in range(s+1,len(n)):
        if n[j]=='(':
            a.append(a[-1]+1)
        elif n[j]==')':
            a.append(a[-1]-1)
    if min(a)<0:
        print(-1)
    else:
        for _ in range(p-1):
            print(1)
        print(b-c-p+1)

