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: Accepted)

#include <iostream>
#include <string>
#include <utility>
#include <vector>
#include <algorithm>

using namespace std;

struct Comp
{
	bool operator()(const pair<int, int>& a, const pair<int, int>& b) const
	{
		if (a.first - a.second <= 0 && b.first - b.second <= 0) {
			return a.first < b.first;
		}
		else if (a.first - a.second > 0 && b.first - b.second > 0) {
			return a.second > b.second;
		}
		return a.first - a.second <= 0;
	}
};

int main()
{
	ios_base::sync_with_stdio(false);
	int n;
	cin >> n;
	int left = 0, right = 0;
	vector<pair<int, int>> mid;
	for (int i = 0; i < n; i++) {
		string t;
		cin >> t;
		int a = 0;
		int c = 0;
		for (int j = 0; j < t.size(); j++) {
			if (t[j] == '(') {
				c++;
			}
			else {
				c--;
				if (c < 0) {
					a++;
					c = 0;
				}
			}
		}
		if (a == 0) {
			left += c;
		}
		else if (c == 0) {
			right += a;
		}
		else {
			mid.push_back({ a, c });
		}
	}

	sort(mid.begin(), mid.end(), Comp());

	int cur = left;
	for (int i = 0; i < mid.size(); i++) {
		cur -= mid[i].first;
		if (cur < 0) {
			cout << "No" << endl;
			return 0;
		}
		cur += mid[i].second;
	}

	if (cur != right) {
		cout << "No" << endl;
		return 0;
	}
	cout << "Yes" << endl;
	return 0;
}