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<iostream>
#include<map>
#include<vector>
#include<algorithm>
#include<cstdlib>
#include<string>
template<typename T>
void fin(T const& t){ std::cout << t << std::endl; exit(0); }

int main() {
  int N; std::cin >> N;
  std::vector<std::pair<int, int>> pls, mis;
  int sum = 0;
  for(int i = 0; i < N; ++i) {
    int mi = 0, cur = 0;
    std::string S; std::cin >> S;
    for(auto s: S) {
      if(s == ')') --cur; else ++cur;
      mi = std::min(mi, cur);
    }
    if(mi != 0 || cur != 0) {
      if(cur >= 0) pls.emplace_back(mi, cur);
      else         mis.emplace_back(mi-cur, -cur);
    }
    sum += cur;
  }
  if(sum != 0) fin("No");

  std::sort(pls.begin(), pls.end(), std::greater<std::pair<int,int>>());
  for(auto p: pls) {
    if(sum + p.first < 0) fin("No");
    sum += p.second;
  }
  sum = 0;
  std::sort(mis.begin(), mis.end(), std::greater<std::pair<int,int>>());
  for(auto p: mis) {
    if(sum + p.first < 0) fin("No");
    sum += p.second;
  }
  fin("Yes");
  return 0;
}
