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>
#define fi first
#define se second
#define pb(x) push_back(x)
#define mp make_pair
#define sz(x) ((int)(x).size())
#define endl "\n"
#define fastio ios_base::sync_with_stdio(false); cin.tie(nullptr);
#define all(x) x.begin(),x.end()
#define y1 omg
#define INF 1000000010
#define FOR(x, n) for(int x=0; x<n; x++)
#define ini(x, n) memset(x, n, sizeof(x));
#define MOD (int)1e9 + 7
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
typedef pair<int, ii> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
#define MD 998244353

int dx[] = { 1, -1, 0, 0, 1, 1, -1, -1 };
int dy[] = { 0, 0, 1, -1, 1, -1, 1, -1 };

int n;
vii v;
string s;

int main() {
	fastio;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> s;
		int a = 0, b = 0;
		for (int i = 0; i < s.size(); i++) {
			if (s[i] == ')') {
				if (b > 0) b--;
				else a++;
			}
			else if (s[i] == '(') {
				b++;
			}
		}
		v.pb(mp(a, -b));
	}
	sort(all(v));
	int cnt = 0;
	for (int i = 0; i < v.size(); i++) {
		cnt -= v[i].fi;
		if (cnt < 0) {
			cout << "No"; 
			return 0;	
		}
		cnt -= v[i].se;
	}
	if (cnt == 0) cout << "Yes";
	else cout << "No";
}