Educational Codeforces Round 45 (Rated for Div. 2)

A bracket sequence is a string containing only characters "(" and ")".

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

You are given nn bracket sequences s1,s2,…,sns1,s2,…,sn . Calculate the number of pairs i,j(1≤i,j≤n)i,j(1≤i,j≤n) such that the bracket sequence si+sjsi+sj is a regular bracket sequence. Operation ++ means concatenation i.e. "()(" + ")()" = "()()()".

If si+sjsi+sj and sj+sisj+si are regular bracket sequences and i≠ji≠j , then both pairs (i,j)(i,j) and (j,i)(j,i) must be counted in the answer. Also, if si+sisi+si is a regular bracket sequence, the pair (i,i)(i,i) must be counted in the answer.

Input

The first line contains one integer n(1≤n≤3⋅105)n(1≤n≤3⋅105) — the number of bracket sequences. The following nn lines contain bracket sequences — non-empty strings consisting only of characters "(" and ")". The sum of lengths of all bracket sequences does not exceed 3⋅1053⋅105 .

Output

In the single line print a single integer — the number of pairs i,j(1≤i,j≤n)i,j(1≤i,j≤n) such that the bracket sequence si+sjsi+sj is a regular bracket sequence.

Examples

Input

3
)
()
(

Output

 

2

Input

 

2
()
()

Output

4

Note

In the first example, suitable pairs are (3,1)(3,1) and (2,2)(2,2) .

In the second example, any pair is suitable, namely (1,1),(1,2),(2,1),(2,2)(1,1),(1,2),(2,1),(2,2) .

题意：有n个字符串，每个字符串都只有'('和')'组成，从中找出两个字符串可以构成完全匹配的个数(这两个字符串也可以由自己本身组成，如(2,2),(1,1).

题解：所有的字符串可以分为3类:1.自身完美匹配型（即左括号和右括号完美匹配）2：除去完全匹配的子串，剩下的都是左括号，3：除去完全匹配的子串，剩下的都是右括号。对于第一类他的个数ans=c(n,2)*A(2,2)+n(它自身构成的完美匹配），对于第二类和第3类，用map查询一遍（如果有左括号的个数等于右括号的个数，ans=(左括号的种类*右括号的种类），最后不要忘记除去2，因为我们算了两遍。还有一点要注意的是一定要用long long ，我错了好几次才发现这一点。

 #include<stdio.h>
 #include<string.h>
 #include<stack>
 #include<string.h>
 #include<queue>
 #include<algorithm>
 #include<iostream>
 #include<map>
 #include<vector>
 #define PI acos(-1.0)
 using namespace std;
 typedef long long ll;
 const int MAXN=3e5+;
 ll  m;
 ll ans;
 char str[MAXN];
 map<ll ,ll>::iterator it;
 int main()
 {
     ll T;
     scanf("%lld",&T);
     map<ll,ll>mp;
     mp.size();
     while(T--)
     {
         stack<char>s;
         scanf(" %s",&str);
         ll len=strlen(str);
         for(ll i=;i<len;i++)
         {
             if(!s.empty())
             {
                 if(s.top()=='('&&str[i]==')')
                 {
                     s.pop();
                 }
                 else
                     s.push(str[i]);
             }
             else
             {
                 s.push(str[i]);
             }
         }
         if(s.empty())
         {
             m++;
         }
         else
         {
             ll cpp=s.size(),flag=;
             while(!s.empty())
             {
                 if(s.top()=='(')
                     flag++;
                 s.pop();
             }
             if(flag==)//栈里面都是右括号
                 mp[-cpp]++;
             else if(flag==cpp)//栈里面都是左括号
             {
                 mp[cpp]++;
             }
         }
     }
     for(it=mp.begin();it!=mp.end();it++)
     {
         ll k=it->first;
         if(mp.count(-k))  {
             ans+=(ll)(it->second*mp[-k]);//左括号的种类*右括号的种类
         }
     }
     printf("%lld\n",ans/+m*m);
 }