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;
const int N=1e6+9;
int n,ac,bc;
struct str{int m;string st;}a[N],b[N];
bool cmp(const str&a,const str&b){return a.m>b.m;}
string last="";
int main(){
	cin>>n;
	for(int i=1;i<=n;i++){
		string s; cin>>s;
		int sum=0,mx=-1e9,mn=1e9;
		for(int j=0;j<s.size();j++){
			if(s[j]=='(') sum++;
			else sum--;
			mn=min(mn,sum);
		} sum=0;
		for(int j=s.size()-1;j>=0;j--){
			if(s[j]=='(') sum++;
			else sum--;
			mx=max(mx,sum);
		}
		if(sum>=0) a[++ac]=(str){mn,s};
		else b[++bc]=(str){mx,s};
	}
	sort(a+1,a+ac+1,cmp), sort(b+1,b+bc+1,cmp);
	for(int i=1;i<=ac;i++) last+=a[i].st;
	for(int i=1;i<=bc;i++) last+=b[i].st;
	int sum=0;
	for(int i=0;i<last.size();i++){
		if(last[i]=='(') sum++;
		else sum--;
		if(sum<0) return puts("No"),0;
	}
	if(sum==0) puts("Yes");
	else puts("No");
	return 0;
}