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> PII;
typedef long long ll;

vector<PII> v;

bool cmp(PII x,PII y) {
	return min(x.second,x.first+y.second)>min(y.second,y.first+x.second);
}

int main() {
	int n;
	cin>>n;
	int sum=0;
	for(int i=1;i<=n;i++) {
		string s;
		int cnt=0,mn=0;//Reason why mn initializes with 0, because cnt should be negative
		cin>>s;
		for(int j=0;j<s.size();j++) {
			if(s[j]=='(') {
				cnt++;
			}else {
				cnt--;
			}
			mn=min(mn,cnt);
		}
		v.push_back(make_pair(cnt,mn));
		sum+=cnt;
	}
	if(sum!=0) {
		cout<<"No";
		return 0;
	}
	sort(v.begin(),v.end(),cmp);
	int s=0;
	for(int i=0;i<v.size();i++) {
		if(s+v[i].second<0) {
			cout<<"No";
			return 0;
		}
		s+=v[i].first;
	}
	cout<<"Yes";
	return 0;
}