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;

void no() {
  cout << "No" << endl;
  exit(0);
}

int main() {
  int n; cin >> n;
  vector<pair<int, int>> op, cl;
  long long total = 0;
  for (int i = 0; i < n; i++) {
    string s; cin >> s;
    int res = 0, dep = 0;
    for (char c : s) {
      res += (c == '(' ? 1 : -1);
      dep = min(res, dep);
    }
    if (res > 0) op.push_back({dep, res});
    else cl.push_back({dep - res, -res});
    total += res;
  }
  if (total != 0) no();
  total = 0;
  sort(op.rbegin(), op.rend());
  for (auto p : op) {
    if (total + p.first < 0) no();
    total += p.second;
  }
  total = 0;
  sort(cl.rbegin(), cl.rend());
  for (auto p : cl) {
    if (total + p.first < 0) no();
    total += p.second;
  }
  cout << "Yes" << endl;
  return 0;
}