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 <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
  int n;
  cin >> n;

  vector<P> l,r;
  rep(ni,n){
    string s;
    cin >> s;
    int h = 0, b = 0;
    rep(i,s.size()){
      if(s[i]=='(') ++h; else --h;
      b = min(b,h);
    }
    if(h>=0) l.emplace_back(b,h); 
    else r.emplace_back(b-h,-h);
  }
  sort(l.rbegin(),l.rend());
  sort(r.rbegin(),r.rend());

  int ls = [&]{
    int t = 0;
    rep(i,l.size()){
      if(t+l[i].first<0) return -1;
      t += l[i].second;
    }
    return t;
  }();
  int rs = [&]{
    int t = 0;
    rep(i,r.size()){
      if(t+r[i].first<0) return -1;
      t += r[i].second;
    }
    return t;
  }();
  if(ls>=0&&rs>=0&&ls==rs){
    cout << "Yes" << endl;
  }else{
    cout << "No" << endl;
  }
}
