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;
  string ss[n];
  for (int i=0; i<n; i++) cin>>ss[i];

  vector<pair<int, int>> pos, neg;
  for (string s: ss) {
    int cnt = 0;
    int min_cnt = 0;
    for (char c: s) {
      if (c == '(') {
        ++cnt;
      } else {
        --cnt;
        min_cnt = min(cnt, min_cnt);
      }
    }
    if (cnt >= 0) {
      pos.push_back(make_pair(min_cnt, cnt));
    } else {
      neg.push_back(make_pair(min_cnt - cnt, -cnt));
    }
  }
  sort(pos.begin(), pos.end());
  reverse(pos.begin(), pos.end());
  int acc1 = 0;
  for (auto p: pos) {
    if (acc1 + p.first < 0) {
      cout<<"No"<<endl;
      return 0;
    }
    acc1 += p.second;
  }
  sort(neg.begin(), neg.end());
  reverse(neg.begin(), neg.end());
  int acc2 = 0;
  for (auto p: neg) {
    if (acc2 + p.first < 0) {
      cout<<"No"<<endl;
      return 0;
    }
    acc2 += p.second;
  }
  if (acc1 == acc2) {
    cout<<"Yes"<<endl;
  } else {
    cout<<"No"<<endl;
  }
}