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> pii;

#define fi first
#define se second
#define mp make_pair
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

const int N = (int)1e6 + 10;

struct Q{
  int A, B;
  bool operator< (const Q &F) const {
    return max(-B,-F.B-A) < max(-F.B,-B-F.A);
  }
};

int main(){
  fastIO;
  int n;
  cin >> n;
  string t;
  int cnt, low;
  vector<Q> exc;
  for(int i = 0; i < n; i ++ ){
    cin >> t;
    cnt = 0;
    low = 0;
    for(auto x : t){
      if(x == '(') cnt ++;
      else cnt -- ;
      low = min(low, cnt);
    }
    exc.push_back({cnt, low});
  }
  sort(exc.begin(), exc.end());
  cnt = 0;
  for(auto x : exc){
    if(cnt + x.B < 0){
      cout << "No\n";
      return 0;
    }
    cnt += x.A;
  }

  if(cnt != 0)
    cout << "No\n";
  else
    cout << "Yes\n";
  return 0;
}