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;
#define ll long long
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
int n;
char s[N];
typedef pair<int, int>pii;
vector<pii>vc;
void cal() {
	int a = 0, b = INF;
	for (int i = 0; s[i]; i++) {
		if (s[i] == '(')a++;
		else a--;
		b = min(b, a);
	}
	vc.push_back(pii(a, b));
}
bool cmp(const pii& a, const pii& b) {
	//if (a.first >= 0 && b.first >= 0)return a.second > b.second;
	if (a.second >= 0 || b.second >= 0) {
		return a.second > b.second;
	}
	return a.first + b.second > b.first + a.second;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%s", s);
		cal();
	}
	//sort(vc.begin(), vc.end(), cmp);
	for (pii p : vc) {
		if (p.second >= 0 && p.first < 0)while (1);
	}
	int ck = 0;
	for (pii p : vc) {
		if (ck + p.second < 0) { puts("No"); return 0; }
		ck += p.first;
	}
	puts(ck ? "No" : "Yes");
	return 0;
}