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;

#define REP(i,n) for(int i = 0; i < (int)(n); i++)

int main(){
  int n;
  cin >> n;
  vector<pair<int, int> > p;
  vector<pair<int, int> > m;
  REP(i, n){
    string s;
    cin >> s;
    int d = 0;
    int mini = 0;
    for(char c: s){
      c == '(' ? ++d : --d;
      if(d < mini) mini = d;
    }
    if(d >= 0){
      p.push_back(make_pair(-mini, d));
    }else{
      m.push_back(make_pair(d - mini, -d));
    }
  }
  sort(p.begin(), p.end());
  sort(m.begin(), m.end());
  int g = 0;
  for(auto& v: p){
    if(g - v.first < 0){
      cout << "No" << endl;
      return 0;
    }
    g += v.second;
  }
  int h = 0;
  for(auto& v: m){
    if(h - v.first < 0){
      cout << "No" << endl;
      return 0;
    }
    h += v.second;
  }
  cout << (g == h ? "Yes" : "No") << endl;
  return 0;
}
