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<iostream>
#include<bits/stdc++.h>
using namespace std;

int n;
vector<pair<int,string> > op,cl,bl;

void process(string s){
	int o = 0,c = 0;
	for(int i=0;i<(int)s.size();i++){
		if(s[i] == '(')o++;
		else c++;
	}
	if(o > c) op.push_back({c,s});
	if(o < c) cl.push_back({o,s});
	if(o == c)bl.push_back({0,s});
}

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	
	cin>>n;
	for(int i=0;i<n;i++){
		string x;
		cin>>x;
		process(x);
	}
	
	sort(op.begin(),op.end());
	sort(cl.begin(),cl.end());
	reverse(cl.begin(),cl.end());
	
	string tot = "";
	for(int i=0;i<(int)op.size();i++){
		tot += op[i].second;
	}
	
	for(int i=0;i<(int)bl.size();i++){
		tot += bl[i].second;
	}
	
	for(int i=0;i<(int)cl.size();i++){
		tot += cl[i].second;
	}
	
	int cnt = 0;
	for(int i=0;i<(int)tot.size();i++){
		if(tot[i] == '(')cnt++;
		else cnt--;
		if(cnt < 0){
			cout<<"No";
			return 0;
		}
	}
	//cout<<tot<<endl;
	//cout<<cnt<<endl;
	if(cnt == 0)cout<<"Yes";
	else cout<<"No";
	return 0;
}
