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 ll long long
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
int n, ck;
char s[N];
typedef pair<int, int>pii;
vector<pii>vc;
void cal() {
	int a = 0, b = INF;
	for (int i = 0; s[i]; i++) {
		if (s[i] == '(')a++;
		else a--;
		b = min(b, a);
	}
	if (b >= 0)ck += a;
	else vc.push_back(pii(a, b));
}
bool cmp(const pii& a, const pii& b) {
	if (a.first >= 0 || b.first >= 0) {
		if (a.first >= 0 && b.first >= 0)return a.second > b.second;
		else return a.first >= 0;
	}
	return a.first + b.second > b.first + a.second;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%s", s);
		cal();
	}
	sort(vc.begin(), vc.end(), cmp);
	for (pii p : vc) {
		if (ck + p.second < 0) { puts("No"); return 0; }
		ck += p.first;
	}
	puts(ck ? "No" : "Yes");
	return 0;
}