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;
bool solve() {
  int N; cin >> N;
  vector<pair<int, int>> A, B;
  for (int i = 0; i < N; i++) {
    string s; cin >> s;
    bool flag = true;
    if (count(s.begin(), s.end(), '(') < count(s.begin(), s.end(), ')')) {
      reverse(s.begin(), s.end());
      for (auto& c : s) c = '(' + ')' - c;
      flag = false;
    }
    int sum = 0, mi = 0;
    for (auto c : s) {
      if (c == '(') sum++;
      else sum--;
      mi = min(mi, sum);
    }
    (flag ? A : B).emplace_back(mi, sum);
  }
  sort(A.rbegin(), A.rend());
  sort(B.rbegin(), B.rend());
  auto f = [](const vector<pair<int, int>>& v) -> int {
    int res = 0;
    for (auto [mi, sum] : v) {
      if (mi + res < 0) return -1;
      res += sum;
    }
    return res;
  };
  int a = f(A);
  int b = f(B);
  if (a < 0 || b < 0 || a != b) return false;
  return true;
}

int main() {
  if (solve()) cout << "Yes" << endl;
  else cout << "No" << endl;
}
