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;
int n;
int dif; 
vector<pair<int,int> > L,R;
bool check(vector<pair<int,int> > v){
	int l=0;
	for(int i=0;i<v.size();++i){
		pair<int,int> p=v[i];
		int dif=l+p.first;
		if(dif<0){
			return 0;
		}
		l+=p.second;
	}
	return 1;
} 
int main(){
	cin>>n;
	for(int i=0;i<n;++i){
		string s;
		cin>>s;
		int l=0,minl=1e9;
		for(int k=0;k<s.size();++k){
			l+=(s[k]=='(')?1:-1;
			minl=min(minl,l);
		}
		dif+=l;
		if(l>0){
			L.push_back({minl,l});
		}
		else{
			R.push_back({minl-l,-l});
		}
	}
	sort(L.rbegin(),L.rend());
	sort(R.rbegin(),R.rend());
	if(dif==0&&check(L)&&check(R)){
		cout<<"Yes\n";
	}
	else{
		cout<<"No\n";
	}
	return 0;
} 