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;
vector<pair<int, int> >pref, suff;
int main(){
	int n; cin>>n;
	for(int i=0; i<n; i++){
		string s="";
		char c=getchar();
		while(c!='(' && c!=')') c=getchar();
		int fi=0, se=0;
		for(;c=='(' || c==')';){
			if(c=='(') se++;
			else{
				if(se==0) fi++;
				else se--;
			}
		//	cout<<c<<" "<<fi<<" "<<se<<endl;
			c=getchar();
		}
	//	cout<<endl;
		if(se>fi) pref.push_back({fi, se});
		else suff.push_back({se, fi});
	}
	sort(pref.begin(), pref.end());
	sort(suff.begin(), suff.end());
	int tot1=0, tot2=0;
	for(pair<int, int>i:pref){
		if(i.first>tot1){
			cout<<"No\n"; return 0;
		}
		tot1+=i.second-i.first;
	}
	for(pair<int, int>i:suff){
		if(i.first>tot2){
			cout<<"No\n"; return 0;
		}
		tot2+=i.second-i.first;
	}
	if(tot1!=tot2){
		cout<<"No\n"; return 0;
	}
	cout<<"Yes\n";
	return 0;
}