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 n;
string s[1000000];
using PP = pair<int, int>;
PP cat(const string& t) {
  int cnt = 0;
  int mn = 0;
  for (char c : t) {
    cnt += (c == '(' ? 1 : -1);
    mn = min(mn, cnt);
  }
  return { -mn, cnt };
}

bool ok(vector<PP> v) {
  sort(v.begin(), v.end());
  int cnt = 0;
  for (auto p : v) {
    int a = p.first;
    int b = p.second;
    if (a > cnt) return false;
    cnt += b;
  }
  return true;
}

bool sub() {
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    for (char c : s[i]) {
      cnt += (c == '(' ? 1 : -1);
    }
  }
  if (cnt != 0) return false;
  
  vector<PP> up;
  vector<PP> dn;
  for (int i = 0; i < n; ++i) {
    auto p = cat(s[i]);
    if (p.second >= 0) {
      up.push_back(p);
    } else {
      dn.push_back({ p.first + p.second, -p.second });
    }
  }
  return ok(up) && ok(dn);
}

int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> s[i];
  cout << (sub() ? "Yes" : "No") << endl;  
}