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.
i = input()

l = len(i)
leach = [0]*l + [0]
hashs = []

for x in range(l):
    if i[x] == '#':
        leach[x+1] = leach[x]
        hashs.append(x+1)
    elif i[x] == '(':
        leach[x+1] = leach[x] + 1
    else:
        leach[x+1] = leach[x] - 1

mins = [10**9] + [0] * (l+1)
for y in range(1, l+1):
    mins[y] = min(mins[y-1], leach[-y])
mins = mins[1:]
mins.reverse()

need = i.count('(') - i.count(')')
if i.count('#') > need:
    print(-1)
else:
    key = 0
    for i in hashs:
        req = mins[i] - key
        if req < need:
            need -= req
            print(req)
            key += req
        elif req >= need:
            print(need)
            quit()
            
