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 long long ll;

struct seg {
  int balance, min;
};
bool operator<(seg a, seg b) {
  return a.min > b.min;
}

bool check(vector<seg> plus) {
  sort(plus.begin(), plus.end());
  int balance = 0;
  for (seg s : plus) {
    if (balance+s.min < 0) return 1;
    balance += s.balance;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  vector<seg> plus, minus;
  int total = 0;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    int balance = 0, mi = 0;
    for (char c : s) {
      balance += (c == '(')-(c == ')');
      mi = min(mi, balance);
    }
    total += balance;
    if (balance >= 0)
      plus.push_back({balance, mi});
    else
      minus.push_back({-balance, mi-balance});
  }
  if (total || check(plus) || check(minus)) {
    cout << "No" << endl;
    return 0;
  }
  cout << "Yes" << endl;
}
