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;
typedef pair<int, int> ii;

const int N = 1e6 + 5;
int n;
vector<ii> p;

bool cmp(ii p1, ii p2) {
  return min(p1.second,p1.first+p2.second)
    > min(p2.second,p2.first+p1.second);
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  cin >> n;
  string s;
  int tt = 0;
  for (int i = 0; i < n; ++i) {
    cin >> s;
    int cnt = 0;
    int mcnt = 0;
    for (auto cc : s) {
      if (cc == '(')
        ++cnt;
      else
        --cnt;
      mcnt = min(mcnt, cnt);
    }
    tt += cnt;
    p.push_back(ii(cnt, mcnt));
  }
  if (tt != 0) {
    cout << "No" << endl;
  } else {
    sort(p.begin(), p.end(), cmp);
    int cnt = 0;
    bool ok = true;
    for (auto it = p.begin(); it != p.end(); ++it) {
      if (it->second + cnt < 0) {
        ok = false;
        break;
      }
      cnt += it->first;
    }
    if (ok)
      cout << "Yes" << endl;
    else
      cout << "No" << endl;
  }
  
  return 0;
}
