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
#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int N=1e6+10;
string str[N];	int n,res;
vector<pair<int,int> > v1,v2;
inline int ch(char c){if(c=='(') return 1; return -1;}
signed main(){
	ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr);
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>str[i];	int s=0,mi=1e9;
		for(int j=0;j<str[i].size();j++)	s+=ch(str[i][j]),mi=min(s,mi);
		if(mi>=0&&s>=0)	res+=s;
		else if(mi<0&&s>=0) v1.push_back({mi,s});
		else v2.push_back({mi,s});
	}
	sort(v1.begin(),v1.end(),[](pair<int,int> a,pair<int,int> b){
		return a.first>b.first;
	});
	for(auto i:v1){
		if(res+i.first<0)	return puts("No"),0;
		res+=i.second;
	}
	sort(v2.begin(),v2.end(),[](pair<int,int> a,pair<int,int> b){
		return a.first-a.second<b.first-b.second;
	});
	for(auto i:v2){
		if(res+i.first<0)	return puts("No"),0;
		res+=i.second;
	}
	puts(res==0?"Yes":"No");
	return 0;
}
