# Treasure

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.

Input

The first line of the input contains a string s (1 ≤ |s| ≤ 105). Each character of this string is one of the characters ‘(‘, ‘)’ or ‘#’. It is guaranteed that s contains at least one ‘#’ character.

Output

If there is no way of replacing ‘#’ characters which leads to a beautiful string print  - 1. Otherwise for each character ‘#’ print a separate line containing a positive integer, the number of ‘)’ characters this character must be replaced with.

If there are several possible answers, you may output any of them.

Examples

input

(((#)((#)

output

12

input

()((#((#(#()

output

221

input

#

output

-1

input

(#)

output

-1

Note

|s| denotes the length of the string s.

Solution:

#include <cstring>
#include <vector>
#include  <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <memory.h>
#include <cassert>

using namespace std;

char s[1234567];

int main() {
scanf("%s", s);
int n = strlen(s);
int total = 0;
int pos = -1;
for (int i = 0; i < n; i++) {
total += (s[i] == '(' ? 1 : -1);
if (s[i] == '#') {
pos = i;
}
}
if (total < 0) {
printf("%d\n", -1);
return 0;
}
int last = 1 + total;
total = 0;
for (int i = 0; i < n; i++) {
total += (i == pos ? -last : (s[i] == '(' ? 1 : -1));
if (total < 0) {
printf("%d\n", -1);
return 0;
}
}
for (int i = 0; i < n; i++) {
if (s[i] == '#') {
if (i == pos) {
printf("%d\n", last);
} else {
printf("%d\n", 1);
}
}
}
return 0;
}