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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

bool check(vector<P> s) {
  int h = 0;
  for (auto p : s) {
      int b = h+p.first;
      if (b < 0) return false;
      h += p.second;
  }
  return true;
}
 
int main() {
  int n;
  cin >> n;
  int total = 0;
  vector<P> ls, rs;
  rep(i,n) {
    int b = 0, h = 0;
    string s;
    cin >> s;
    for (char c : s) {
      if (c == '(') ++h; else --h;
      b = min(b,h);
    }
    if (h > 0) ls.emplace_back(b,h);
    else rs.emplace_back(b-h,-h);
    total += h;
  }
  sort(ls.rbegin(), ls.rend());
  sort(rs.rbegin(), rs.rend());
  if (check(ls) && check(rs) && total == 0) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
  return 0;
}