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;

typedef long long ll;

ll N;
typedef pair<ll, ll> Pair;
priority_queue<Pair> Asc, Desc;

bool solve() {
  ll cnt = 0;
  while (!Asc.empty()) {
    ll bot, diff;
    tie(bot, diff) = Asc.top();
    Asc.pop();
    if (bot + cnt < 0) return false;
    cnt += diff;
  }
  ll cnt2 = 0;
  while (!Desc.empty()) {
    ll bot, diff;
    tie(bot, diff) = Desc.top();
    Desc.pop();
    if (bot + cnt2 < 0) return false;
    cnt2 += diff;
  }
  return cnt == cnt2;
}
int main() {
  cin >> N;
  for (ll i = 0; i < N; ++i) {
    string s;
    cin >> s;
    ll cnt = 0, bot = 0;
    for (ll j = 0; j < s.size(); ++j) {
      if (s[j] == '(') {
        ++cnt;
      } else {
        --cnt;
        bot = min(bot, cnt);
      }
    }
    if (cnt > 0)
      Asc.push({bot, cnt});
    else
      Desc.push({bot - cnt, -cnt});
  }
  cout << (solve() ? "Yes" : "No") << endl;
  return 0;
}
