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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <cmath>
#include <stdio.h>
#include <queue>
#include <deque>
#include <cstdio>
#include <set>
#include <map>
#include <bitset>
#include <stack>
#include <cctype>
using namespace std;
long long g[1010000] = { 0 };
int main() {
#define int long long
	int n;
	vector<int> l, r;
	string s;
	cin >> n;
	for (int j = 0; j < n; j++) {
		cin >> s;
		int l1 = 0, r1 = 0;
		for (int i = 0; i < s.size(); i++) {
			if (s[i] == ')') {
				if (l1 != 0) {
					l1--;
				}
				else {
					r1++;
				}
			}
			else {
				l1++;
			}
		}
		l.push_back(l1);
		r.push_back(r1);
		if (s.size() != 1) {
			if (s[0] == ')') {
				if (s[s.size() - 1] == '(') {
					g[j] = 1;
				}
			}
		}

	}
	int x = 0, y = 0;
	bool t = false, t1 = false;
	int co3 = 0;
	for (int i = 0; i < n; i++) {
		if (l[i] != 0 || r[i] != 0) {
			if (g[i] == 0) {
				if (l[i] != 0) {
					t = true;
				}
				if (r[i] != 0) {
					t1 = true;
				}
			}
		}
		else {
			co3++;
		}
		x += l[i];
		y += r[i];
	}
	if (t == true && t1 == true) {
		if (x - y == 0) {
			cout << "Yes" << endl;
			return 0;
		}
	}
	if (co3 == n) {
		if (x - y == 0) {
			cout << "Yes" << endl;
			return 0;
		}
	}
	cout << "No" << endl;
	return 0;
}