990C. Bracket Sequences Concatenation Problem（思维+括号匹配问题）

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 $n$

bracket sequences $s_1,s_2,\dots ,s_n$. Calculate the number of pairs $i,j(1\le i,j\le n)$ such that the bracket sequence $s_i+s_j$ is a regular bracket sequence. Operation $+$

means concatenation i.e. "()(" + ")()" = "()()()".

If $s_i+s_j$

and $s_j+s_i$ are regular bracket sequences and $i\ne j$, then both pairs $(i,j)$ and $(j,i)$ must be counted in the answer. Also, if $s_i+s_i$ is a regular bracket sequence, the pair $(i,i)$

must be counted in the answer.

Input

The first line contains one integer $n(1\le n\le 3\cdot {10}^5)$

— the number of bracket sequences. The following $n$ lines contain bracket sequences — non-empty strings consisting only of characters "(" and ")". The sum of lengths of all bracket sequences does not exceed $3\cdot {10}^5$

.

Output

In the single line print a single integer — the number of pairs $i,j(1\le i,j\le n)$

such that the bracket sequence $s_i+s_j$

is a regular bracket sequence.

Examples

Input

Copy

3
)
()
(

Output

Copy

2

Input

Copy

2
()
()

Output

Copy

4

Note

In the first example, suitable pairs are $(3,1)$

and $(2,2)$

.

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

.

题意：有n个字符串，每个字符串只由‘’（‘’，‘’）‘’构成，从n个字符串中任意选两个a和b，如果ab构成的字符串括号完全匹配，则结果+1，也可以是aa，ba结构，求最后的结果

思路：括号匹配问题

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 3e5+5;
string str[N];
int vis[N];
int solve(string s){
	int x=0;
	for(int i=0;i<s.length();i++){
		if(s[i]=='(') x++;
		else x--;
		if(x<0) return -1;
	}
	return x;
}
string rev(string s){
	reverse(s.begin(),s.end());
	for(int i=0;i<s.length();i++){
		if(s[i]=='(') s[i]=')';
		else s[i]='(';
	}
	return s;
}
int main(){
	int n;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>str[i];
		int x=solve(str[i]);
		if(x!=-1) vis[x]++;
	}
	LL ans=0;
	for(int i=0;i<n;i++){
		str[i]=rev(str[i]);
		int x=solve(str[i]);
		if(x!=-1) ans+=vis[x];
	}
	cout<<ans<<endl;
	return 0;
}