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 pii pair<int,int>
using namespace std;
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin >> n;
	int cnt[2] = {};
	vector<pii> arr;
	while(n--) {
		string s;
		cin >> s;
		
		int a = 0 , b = 0;
		for(char c : s)
			if(c == '(') {
				a++;
			}
			else {
				if(a == 0)
					b++;
				else
					a--;
			}
		
		if(a == 0 && b == 0)	continue;
		if(a == 0)
			cnt[1] += b;
		else if(b == 0)
			cnt[0] += a;
		else 
			arr.push_back({a,b});
	}
	sort(arr.begin(),arr.end(),[=](pii a,pii b)->bool
		{return (a.first > a.second && b.first > b.second) ? a.second < b.second : a.first - a.second > b.first - b.second;});
	for(pii now : arr) {
		cnt[0] -= now.second;
		if(cnt[0] < 0) {
			cout << "No" << endl;
			return 0;
		}
		cnt[0] += now.first;
	}
	if(cnt[0] == cnt[1])
		cout << "Yes" << endl;
	else
		cout << "No" << endl;
	return 0;
}