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;

int main() {
  int n;
  cin >> n;
  vector<string> s(n);
  for (int i = 0; i < n; i++) {
    cin >> s.at(i);
  }
  priority_queue<pair<int, int>> q0, q1;
  for (int i = 0; i < n; i++) {
    int m = 0, t = 0;
    for (char c : s.at(i)) {
      if (c == '(') {
        t++;
      } else {
        t--;
        m = min(m, t);
      }
    }
    if (t >= 0) {
      q0.emplace(m, t);
    } else {
      q1.emplace(m - t, -t);
    }
  }
  int k0 = 0, k1 = 0;
  while (!q0.empty()) {
    auto x = q0.top();
    q0.pop();
    int m = x.first, t = x.second;
    if (k0 + m < 0) {
      cout << "No" << endl;
      return 0;
    }
    k0 += t;
  }
  while (!q1.empty()) {
    auto x = q1.top();
    q1.pop();
    int m = x.first, t = x.second;
    if (k1 + m < 0) {
      cout << "No" << endl;
      return 0;
    }
    k1 += t;
  }
  if (k0 == k1) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
}
