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<iostream>
#include<vector>
#include<string>
#include<utility>
#include<algorithm>

bool check(std::vector<std::pair<int, int> > x){
	int y = 0, low;
	for(std::pair<int, int> p : x){
		low = y + p.first;
		if(low < 0){
			return false;
		}
		y += p.second;
	}
	return true;
}

int main(){
	int n;
	int total = 0;
	std::cin >> n;
	std::vector<std::pair<int, int> > left, right;
	//<lowest, y>
	//lowest = 最も低くなる位置
	//y = 最終的な位置

	for(int i=0; i<n; i++){
		std::string s;
		std::cin >> s;
		int y = 0, lowest = 0;
		for(char j : s){
			if(j == '('){
				y++;
			}else{
				y--;
			}
			lowest = std::min(y, lowest);
		}

		if(y > 0){
			left.emplace_back(lowest, y);
		}else{
			right.emplace_back(lowest - y, -y);
		}
		total += y;
	}

	//lowestが大きいものを優先する
	sort(left.rbegin(), left.rend());
	sort(right.rbegin(), right.rend());
	if(check(left) && check(right) && total == 0){
		std::cout << "Yes" << std::endl;
	}else{
		std::cout << "No" << std::endl;
	}
	return 0;
}