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;
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int type_cnt[2] = {} , l = 0 , r = 0;
	int n;
	cin >> n;
	while(n--) {
		string s;
		cin >> s;
		
		int a = 0 , b = 0;
		for(char c : s)
			if(c == '(') {
				a++;
			}
			else {
				if(a == 0)
					b++;
				else
					a--;
			}
		
		l += a;		r += b;
		if(a == 0 && b == 0)	continue;
		if(a == 0)
			type_cnt[0]++;
		else if(b == 0)
			type_cnt[1]++;
	}
	if((l == 0 && r == 0) || (type_cnt[0] > 0 && type_cnt[1] > 0 && l == r))
		cout << "Yes" << endl;
	else
		cout << "No" << endl;
	return 0;
}