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;
using P = pair<int,int>;

int main(){
  int n;
  cin >> n;
  priority_queue<P> que1, que2;
  for(int i=0; i<n; ++i){
    string s;
    cin >> s;
    int sz = s.size(), cnt=0, mincnt = 0;
    for(int j=0; j<sz; ++j){
      if(s[j] == '(') ++cnt;
      else{
        --cnt;
        mincnt = min(mincnt, cnt);
      }
    }
    if(cnt >= 0) que1.push(P(mincnt,cnt));
    else que2.push(P(mincnt-cnt, -cnt));
  }
  int t1 = 0;
  while(!que1.empty()){
    int x, y;
    tie(x,y) = que1.top();
    que1.pop();
    if(x+t1 <0){
      cout << "No" << endl;
      return 0;
    }
    t1 += y;
  }
  int t2 = 0;
  while(!que2.empty()){
    int x, y;
    tie(x,y) = que2.top();
    que2.pop();
    if(x+t2 <0){
      cout << "No" << endl;
      return 0;
    }
    t2 += y;
  }
  cout << (t1 == t2 ? "Yes":"No") << endl;
}