Problem:
Score : 600 points

Problem Statement

A bracket sequence is a string that is one of the following:

An empty string;

The concatenation of (, A, and ) in this order, for some bracket sequence A ;

The concatenation of A and B in this order, for some non-empty bracket sequences A and B /

Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?

Constraints

1 \leq N \leq 10^6

The total length of the strings S_i is at most 10^6.

S_i is a non-empty string consisting of ( and ).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

If a bracket sequence can be formed by concatenating all the N strings in some order, print Yes; otherwise, print No.

Sample Input 1

2
)
(()

Sample Output 1

Yes

Concatenating (() and ) in this order forms a bracket sequence.

Sample Input 2

2
)(
()

Sample Output 2

No

Sample Input 3

4
((()))
((((((
))))))
()()()

Sample Output 3

Yes

Sample Input 4

3
(((
)
)

Sample Output 4

No

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std ;

const int MAX = 1e6 + 10 ;

int arr[MAX] , Plus[MAX] , Minus[MAX] ;
int n ;

bool cmp(int &i , int &j)
{
	if(Minus[i] == 0 && Minus[j] == 0)
		return Plus[i] > Plus[j] ;
	else if(Minus[i] == 0 || Minus[j] == 0) // return one the zero one
		return Minus[i] < Minus[j] ;
	else
		return Plus[i] > Plus[j] ; 
}

int main()
{
	ios_base::sync_with_stdio(0) ;
	cin.tie(0) ;
	cin>>n ;
	string s ;
	vector<int>perm ;
	for(int i = 0 ; i < n ; ++i)
	{
		cin>>s ;
		int cnt = 0 ;
		for(auto &c : s)
		{
			if(c == '(')
				cnt++ ;
			else
				cnt-- ;
			if(cnt < 0)
				Minus[i]++ ;
		}
		if(cnt > 0)
			Plus[i] = cnt ;
		perm.push_back(i) ;
	}
	sort(perm.begin() , perm.end() , cmp) ;
	int cnt = 0 ;
	for(auto &i : perm)
	{
		cnt -= Minus[i] ;
		if(cnt < 0)
			return cout<<"No\n" , 0 ;
		cnt += Plus[i] ;
	}
	if(cnt == 0)
		cout<<"Yes\n" ;
	else
		cout<<"No\n" ;
	return 0 ;
}