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;

long long f(priority_queue<pair<int, int>>& pq){
  long long sum = 0;
  while(!pq.empty() && sum + pq.top().first >= 0){
    sum += pq.top().second;
    pq.pop();
  }
  
  if(pq.empty()) return sum;
  else return -1;
}

int main(){
  int N;
  cin >> N;
  
  priority_queue<pair<int, int>> pqa, pqd;
  for(int i = 0; i < N; i++){
    string S;
    cin >> S;
    int b = 0, c = 0;
    for(char ch : S){
      if(ch == '(') ++c;
      else --c;
      b = min(c, b);
    }
    
    if(c >= 0) pqa.push(make_pair(b, c));
    else pqd.push(make_pair(b - c, -c));
  }
  
  long long suma = f(pqa), sumd = f(pqd);
  if(suma >= 0 && suma == sumd) puts("Yes");
  else puts("No");
  
  return 0;
}