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;
#define ff first
#define ss second
#define len(x) (int)(x.size())
#define all(vs) vs.begin(), vs.end()
using ii = pair<int, int>;
using ll = long long;

char s[1000010];

bool comp(ii a, ii b){
	return a.ss > b.ss;
}
int32_t main(){
	int n;
	scanf("%d", &n);
	vector<ii> v1, v2;
	for(int z=0; z<n; ++z){
		scanf("%s", s);
		int tam = int(strlen(s));
		int q = 0;
		int a = 0, b = 0;
		for(int i=0; i<tam; ++i){
			if(s[i]=='(') ++q;
			else{
				if(q==0) ++b;
				else --q;
			}
		}
		a = q;
		if(b>=a) v1.push_back(ii(a, b));
		else v2.push_back(ii(a, b));
	}
	sort(all(v1));
	sort(all(v2), comp);
	int r = 0;
	for(auto x:v1){
		int a = x.ff;
		int b = x.ss;
		if(a>r){ 
			printf("No\n");
			return 0;
		}
		r-=a;
		r+=b;
	}
	for(auto x:v2){
		int a = x.ff;
		int b = x.ss;
		if(a>r){ 
			printf("No\n");
			return 0;
		}
		r-=a;
		r+=b;
	}
	if(r==0)
		printf("Yes\n");
	else
		printf("No\n");
}
