cf3D Least Cost Bracket Sequence

This is yet another problem on regular bracket sequences.

A bracket sequence is called regular, if by inserting "+" and "1" into it we get a correct mathematical expression. For example, sequences "(())()", "()" and "(()(()))" are regular, while ")(", "(()" and "(()))(" are not. You have a pattern of a bracket sequence that consists of characters "(", ")" and "?". You have to replace each character "?" with a bracket so, that you get a regular bracket sequence.

For each character "?" the cost of its replacement with "(" and ")" is given. Among all the possible variants your should choose the cheapest.

Input

The first line contains a non-empty pattern of even length, consisting of characters "(", ")" and "?". Its length doesn't exceed 5·10⁴. Then there follow m lines, where m is the number of characters "?" in the pattern. Each line contains two integer numbers a_i and b_i (1 ≤ a_i, b_i ≤ 10⁶), where a_i is the cost of replacing the i-th character "?" with an opening bracket, and b_i — with a closing one.

Output

Print the cost of the optimal regular bracket sequence in the first line, and the required sequence in the second.

Print -1, if there is no answer. If the answer is not unique, print any of them.

Example

Input

(??)
1 2
2 8

Output

4
()()

给一个包含'(' ')' '?'的括号序列，其中'?'需要替换成'(' ')'使得它变成一个合法的括号序列，每个'?'都有一个变成‘(’ ')'的费用，要使得费用最小

不妨先把所有'?'用')'替换，然后再考虑把其中的一些')'改成'('。这样改回'('的费用就是ai-bi

如果'?'改成了')'导致前k个字符不是合法的括号序列了，就把前k个字符中找个ai-bi最小的改掉，这个维护个堆就好

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<deque>
 9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<int,int>
15 #define mkp(a,b) make_pair(a,b)
16 #define pi 3.1415926535897932384626433832795028841971
17 using namespace std;
18 inline LL read()
19 {
20     LL x=0,f=1;char ch=getchar();
21     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
22     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
23     return x*f;
24 }
25 priority_queue<pa,vector<pa>,greater<pa>  >q;
26 char s[100010];
27 int pos[100010];
28 int l[100010],r[100010];
29 int n,p;
30 LL ans;
31 int main()
32 {
33     scanf("%s",s+1);n=strlen(s+1);
34     for (int i=1;i<=n;i++)
35     {
36         if (s[i]=='(')p++;
37         else if (s[i]==')')
38         {
39             p--;
40             if (p<0)
41             {
42                 if (q.empty()){puts("-1");return 0;}
43                 p+=2;
44                 pa now=q.top();q.pop();
45                 ans+=now.first;s[now.second]='(';
46             }
47         }else if (s[i]=='?')
48         {
49             int l=read(),r=read();
50             s[i]=')';p--;
51             ans+=r;q.push(mkp(l-r,i));
52             if (p<0)
53             {
54                 if (q.empty()){puts("-1");return 0;}
55                 p+=2;
56                 pa now=q.top();q.pop();
57                 ans+=now.first;s[now.second]='(';
58             }
59         }
60     }
61     while (p<0)
62     {
63                 p+=2;
64                 pa now=q.top();q.pop();
65                 ans+=now.first;s[now.second]='(';
66     }
67     p=0;
68     for (int i=1;i<=n;i++)
69     {
70         if (s[i]=='(')p++;else p--;
71         if (p<0){puts("-1");return 0;}
72     }
73     if (p!=0){puts("-1");return 0;}
74     printf("%lld
",ans);
75     puts(s+1);
76 }

cf 3D