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;
typedef pair<int, int> pii;
#define mp(a, b) make_pair(a, b)
const int maxn = 1e6 + 10;
pii a[maxn];
vector<pii> b, c;
bool cmp1(const pii & x, const pii & y) { return x.first < y.first;}
bool cmp2(const pii & x, const pii & y) { return x.second > y.second;} 
int main()
{
	int n;
	string s;
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i)
	{
		cin >> s;
		for(auto v : s)
		{
			if(v == '(')
				a[i].second++;
			else
			{
				if(a[i].second)
					--a[i].second;
				else
					++a[i].first;
			}
		}
		if(a[i].second >= a[i].first)
			b.push_back(a[i]);
		else 
			c.push_back(a[i]);
	}
	sort(b.begin(), b.end(), cmp1);
	sort(c.begin(), c.end(), cmp2);
	pii cur = mp(0, 0);
	for(auto v : c) b.push_back(v);
	for(auto v : b)
	{
		int e = min(cur.second, v.first);
		cur.second -= e;
		v.first -= e;
		cur.first += v.first;
		cur.second += v.second;
	}
	if(cur.first == 0 && cur.second == 0)
		puts("Yes");
	else
		puts("No");
	return 0;
}
/*
5
(()
(()
)))(
(
)
*/