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 <bits/stdc++.h>

#define endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define forn(i,n) for (int i = 0; i < n; i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb(x) push_back(x)

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> ii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ii> vii;

const int MX = 1000005;
string s[MX];
int n;
ii a[MX];
vector<string> f, g;

void no () {
	cout << "No" << endl;
	exit(0);
}

int solve (vector<string> &s) {
	int n = s.size();

	forn (i, n) {
		int x = 0, mn = 0;
	
		for (char c : s[i]) {
			if (c == '(') x++;
			else x--;

			mn = min(mn, x);
		}

		a[i] = {mn, x};
	}

	sort(a, a + n, greater<ii>());

	int sum = 0;
	forn (i, n) {
		if (sum + a[i].fi < 0) no();
		sum += a[i].se;
	}

	return sum;
}

int main () {
	ios_base::sync_with_stdio(0); cin.tie(0);

	cin >> n;

	forn (i, n) {
		cin >> s[i];

		int x = 0;

		for (char c : s[i]) {
			if (c == '(') x++;
			else x--;
		}

		if (x > 0) f.pb(s[i]);
		else {
			reverse(all(s[i]));
			for (char &c : s[i]) {
				if (c == '(') c = ')';
				else c = '(';
			}
			g.pb(s[i]);
		}
	}

	if (solve(f) == solve(g)) cout << "Yes" << endl;
	else cout << "No" << endl;

	return 0;
}