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| ≤ 105). Each character of this string is one of the characters '(', ')' or '#'. It is guaranteed that s contains at least one '#' character.

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.

|s| denotes the length of the string s.

题意是要把#替换成1个以上的")"，使得"("和“)”的个数相等，且对于s的任意一个前缀，“)“的个数不大于”)“的个数。

因为个数相等，所以分配给#的”)“的个数之和是确定的。

然后很显然的贪心是前k-1个#都只分配一个")"，最后一个#多分配一点")"使”(“和”)"的个数相等就好了

于是变成模拟题了

#include<cstdio>

#include<iostream>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<queue>

#include<deque>

#include<set>

#include<map>

#include<ctime>

#define LL long long

#define inf 0x7ffffff

#define pa pair<int,int>

#define pi 3.1415926535897932384626433832795028841971

using namespace std;

inline LL read()

{

    LL x=0,f=1;char ch=getchar();

    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

    return x*f;

}

inline void write(LL a)

{

	if (a<0){printf("-");a=-a;}

	if (a>=10)write(a/10);

	putchar(a%10+'0');

}

char ch[100010];

int a[100010];

int s[100010];

int len,x,y,z,lst;

int main()

{

	scanf("%s",ch);

	for (int i=1;ch[i-1];i++)

	{

		if (ch[i-1]=='(')a[i]=1,x++;

		if (ch[i-1]==')')a[i]=-1,y++;

		if (ch[i-1]=='#')a[i]=0,z++,lst=i;

		len=i;

	}

	if (x-y<z)

	{

		printf("-1");

		return 0;

	}

	for (int i=1;i<=len;i++)

	{

		s[i]=s[i-1];

		if (a[i]!=0)s[i]+=a[i];

		else if (i==lst)s[i]-=x-(y+z-1);

		else s[i]--;

		if (s[i]<0)

		{

			printf("-1");

			return 0;

		}

	}

	for (int i=1;i<z;i++)

	  printf("1\n");

	printf("%d\n",x-y-(z-1));

}