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 <iostream>
#include <vector>
#include <utility>
#include <algorithm>

using namespace std;

typedef pair<int,int> ii;
vector <ii> su,r;

int main(){
	
	cin.tie(0); ios_base::sync_with_stdio(0);
	string s;
	int n;
	ii p;
	cin>>n;
	while(n--){
		cin>>s;
		p.first = p.second =0;
		for(int i=0;i<s.length();i++){
			if(s[i]=='(')p.second++;
			else{
				if(p.second)p.second--;
				else p.first++;
			}
		}
		if(p.second>p.first)su.push_back(p);
		else r.push_back(ii(p.second,p.first));
	}
	sort(su.begin(),su.end());
	sort(r.begin(),r.end());
	int o = 0;
	for(int i=0;i<su.size();i++){
		if(su[i].first>o){
			cout<<"No\n";
			return 0;
		}
		o+=su[i].second-su[i].first;
	}
	int o2 =0;
	for(int i=0;i<r.size();i++){
		if(r[i].first>o2){
			cout<<"No\n";
			return 0;
		}
		o2+=r[i].second-r[i].first;
	}
	if(o!=o2){
		cout<<"No\n";
		return 0;
	}
	cout<<"Yes\n";
	
	return 0;
}