CodeForces 990C

Description

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.

Sample Input

Input

3
)
()
(

Output

2

Input

2
()
()

Output

4

Hint

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).

括号匹配问题，需要注意几个细节。

有以下几种情况是不能与其他组括号构成合法的括号：)(         )(((      )))(

已经是合法的括号组不需要与不合法的括号组匹配。

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #include<stack>
 #include<deque>
 #include<map>
 #include<iostream>
 using namespace std;
 typedef long long  LL;
 const double pi=acos(-1.0);
 const double e=exp();
 const int N = ;

 #define lson i << 1,l,m
 #define rson i << 1 | 1,m + 1,r

 map<LL,LL> mp;
 char con[N];
 int main()
 {
     LL i,p,j,n,check;
     LL cont=,ans=,len1,len2;
     scanf("%lld",&n);
     getchar();
     for(j=;j<=n;j++)
     {
         p=check=;
         len1=len2=;
         memset(con,,sizeof());
         scanf("%s",con);
         for(i=;i<;i++)
         {
             if(con[i]==)
                 break;
             if(con[i]=='(')
             {
                 len1++;
                 p++;
             }
             else
             {
                 p--;
                 if(len1)
                     len1--;
                 else
                     len2++;
             }
         }
         if(len1==&&len2==)
             cont++;
         else
         {
             if(len1==)
                 mp[p]++;
             if(len2==)
                 mp[p]++;
         }
     }
     ans=cont*cont;

     map<LL,LL>::iterator it1;
     for(it1=mp.begin();it1!=mp.end();it1++)
     {
         if(it1->first>)
             break;
         if(mp[-(it1->first)]>)
             ans+=(it1->second)*mp[-(it1->first)];
     }
     printf("%lld\n",ans);
     return ;
 }