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>
#define pb push_back
#define mk make_pair
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair <int,int> PAIR;
const int N = 2e6+5;
const ll M = 998244353;
struct node{
	int x,y;
}a[N];
bool cmp(node p1,node p2){
	return p1.x-p1.y>p2.x-p2.y;
}
int main() {
	int n;
	string s;
	cin>>n;
	getchar();
	int le=0,ri=0;
	for(int i=1;i<=n;i++){
		getline(cin,s);
		int l=0,r = 0;
		for(int j=0;j<s.length();j++){
			if(s[j]==')'){
				if(l>0) l--;
				else r++;
			}else {
				l++;
			}
		}
		if(l&&!r) le+=l;
		else if(r&&!l) ri+=r;
		else a[i].x = l,a[i].y=r;
	}
	sort(a+1,a+1+n,cmp);
	int l = le,r = ri;
	for(int i=1;i<=n;i++){
		if(l<a[i].y){
			cout<<"No\n";
			return 0;
		}else {
			l-=a[i].y;
			l+=a[i].x;
		}
	}
	if(l!=r) cout<<"No\n";
	else cout<<"Yes\n";
	return 0;
}
