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 <string>
#include <vector>
#include <utility>
#include <algorithm>
#include <cstdio>
using namespace std;

typedef pair<int,int> pii;

int func(vector<pii> v){
	sort(v.begin(), v.end());
	int s = 0;
	for(size_t i = 0; i < v.size(); ++i){
		if(s < v[i].first){ return -1; }
		s += v[i].second;
	}
	return s;
}

bool solve(){
	int n;
	cin >> n;
	vector<string> s(n);
	for(int i = 0; i < n; ++i){
		cin >> s[i];
	}
	
	vector<pii> pos, neg;
	for(int i = 0; i < n; ++i){
		int c = 0, m = 0;
		for(size_t j = 0; j < s[i].size(); ++j){
			if(s[i][j] == '('){
				++c;
			}
			else{
				--c;
				m = min(m, c);
			}
		}
		
		if(c >= 0){
			pos.push_back(pii(-m, c));
		}
		else{
			neg.push_back(pii(c - m, -c));
		}
	}

	int a = func(pos), b = func(neg);
	return a >= 0 && a == b;
}

int main(){
	puts(solve() ? "Yes" : "No");
}
