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 long long ll;
const ll mod=1e9+7;
const ll N=1e5+1;
const double PI=acos(-1.0);

bool comp1(pair<ll,ll> a,pair<ll,ll> b){
	//if(a.second==b.second) return a.first>b.first; 
	return a.second<b.second;
}
bool comp2(pair<ll,ll> a,pair<ll,ll> b){
	//if(a.first==b.first) return a.second<b.second;
	return a.first>b.first;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);
	int n;
	cin>>n;
	vector<pair<ll,ll>> pl,pr;
	for(int i=0;i<n;i++){
		string s;
		cin>>s;
		int a=0,b=0;
		for(char ch:s) ch=='('?a++:(a?a--:b++);
		a>=b?pl.push_back({a,b}):pr.push_back({a,b});
	}
	sort(pl.begin(),pl.end(),comp1);
	sort(pr.begin(),pr.end(),comp2);
	for(auto i:pr) pl.push_back(i);
	int sum=0,ans=1;
	for(int i=0;i<pl.size();i++){
		sum-=pl[i].second;
		if(sum<0){
			ans=false;
			break;
		}
		sum+=pl[i].first;
	}
	if(ans and sum) ans=false;
	cout<<(ans?"Yes":"No");
}
