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;
#define int long long 
#define pi pair<int,int>
#define ff first
#define ss second
const int N = 1e5 + 5;


int32_t main() { 
	int n;
	cin>>n;
	string s;
	vector<pi>v;
	int cnt = 0,c = 0;
	for(int i=0;i<n;i++){
		cin>>s;
		int open = 0,close = 0;
		for(char &c:s){
			if(c == '(')
				open++;
			else{
				if(open>0)
					open--;
				else
					close++;
			}
		}
		if(open>0 && close>0)
			v.push_back({open,close});
		else if(open>0)
			cnt+=open;
		else
			c+=close;
	}

	sort(v.begin(),v.end(),[&](pi a,pi b){
		return (a.ff-a.ss > b.ff-b.ss);
	});
	for(pi &p:v){
		cnt-=p.ss;
		if(cnt<0)
			break;
		cnt+=p.ff;
	}
	cnt -=c;
	if(cnt == 0)
		cout<<"Yes"<<endl;
	else
		cout<<"No"<<endl;
}	

