A bracket sequence is a string that is one of the following:

1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. 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`.

Examples

Input

2
)
(()


Output

Yes


Input

2
)(
()


Output

No


Input

4
((()))
((((((
))))))
()()()


Output

Yes


Input

3
(((
)
)


Output

No
#include <bits/stdc++.h>
 
using namespace std;
 
typedef pair<int,int> pi;
typedef pair<pi,string> pis;
 
int N;
vector<pi> good, bad;
string s;
 
bool cmp(pi A, pi B){
	return A.second + A.first > B.second + B.first;
}
 
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie();
	cin >> N;
	for (int i = 1; i <= N; ++i){
		cin >> s;
		int op = 0, clo = 0, dep = 0;
		for (auto it : s){
			((it == '(') ? op++ : clo++);
			dep = max(dep,clo-op);
		}
		if (op >= clo){
			good.push_back(pi(dep,op-clo));
		}
		else bad.push_back(pi(dep,op-clo));
	}
	sort(good.begin(),good.end());
	sort(bad.begin(),bad.end(),cmp);
	int num = 0;
	for (auto it : good){
		if (num < it.first){
			cout << "No\n";
			return 0;
		}
		num += it.second;
	}
	for (auto it : bad){	
		if (num < it.first){
			cout << "No\n";
			return 0;
		}
		num += it.second;
	}
	if (num == 0){
		cout << "Yes\n";
	}
	else cout << "No\n";
}