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<cmath>
#include <iomanip>
#include<string>
#include<algorithm>
#include<vector>

using namespace std;
using LL = long long;

LL n,l,r,R;

string s[1000000],S;

int main() {

	cin >> n;

	for (LL i = 0; i < n; i++) {
		cin >> s[i];
	}

	for (LL i = 0; i < n; i++) {
		for (LL j = 0; j < s[i].size(); j++) {
			if (s[i][j] == '(')l++;
			else r++;
		}
	}
	if (l != r) {
		cout << "No" << endl;
		return 0;
	}

	for (LL i = 0; i < n; i++) {
		for (LL j = 0; j < s[i].size()-1; j++) {
			if (s[i][j] == '(' && s[i][j + 1] == ')') {
				s[i].replace(j, 2, "");
				j=-1;
				if (s[i].size() <= 1)goto A;
			}
		}
	A:;
	}

	for (LL i = 0; i < n; i++) {
		if (!s[i].empty()) {
			if (s[i][0] == '(' && s[i][s[i].size()-1] == '(') {
				S += s[i];
				s[i]="";
			}
		}
	}

	for (LL i = 0; i < n; i++) {
		if (!s[i].empty()) {
			if (s[i][0] == ')' && s[i][s[i].size()-1] == '(') {
				S += s[i];
				s[i] = "";
			}
		}
	}

	for (LL i = 0; i < n; i++) {
		if (!s[i].empty()) {
				S += s[i];
		}
	}

	for (LL i = 0; i < S.size();i++) {
		if (S[i] == '(')R++;
		if (S[i] == ')') R--;
		if (R < 0) {
			cout << "No" << endl;
			return 0;
		}
	}

	cout << "Yes" << endl;

	return 0;
}