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;
bool cmp(pair<int, int>a, pair<int, int>b){
	return a.first+a.second>b.first+b.second;
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int n;
	cin>>n;
	vector<pair<int, int>>v1, v2;
	int tot=0;
	for(int i=0;i<n;i++){
		string s;
		cin>>s;
		int cnt=0, mini=0;
		for(int j=0;j<s.size();j++){
			if(s[j]=='(') cnt++;
			else cnt--;
			mini=min(mini, cnt);
		}
		tot+=cnt;
		if(cnt>=0) v1.push_back({-mini, cnt});
		else v2.push_back({-mini, cnt});
	}
	if(tot!=0){
		cout<<"No";
		return 0;
	}
	sort(v1.begin(), v1.end());
	sort(v2.begin(), v2.end(), cmp);
	int cur=0;
	for(int i=0;i<v1.size();i++){
		if(v1[i].first>cur){
			cout<<"No";
			return 0;
		}
		cur+=v1[i].second;
	}
	//cout<<v1.size()<<endl;
	for(int i=0;i<v2.size();i++){
		if(v2[i].first>cur){
			cout<<"No";
			return 0;
		}
		cur+=v2[i].second;
	}
	cout<<"Yes";
}