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;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;

int main()
{
ll n ;
cin >> n ;
vector<pair<ll,ll>> v,vv;
ll cnt=0;
rep(i,n){
  string s ;
  cin >> s ;
  ll te=0,tt=0;
  rep(j,s.size()){
    if(s[j]==')'&&te==0){tt++;cnt--;}
    else if(s[j]==')'){te--;cnt--;}
    else {te++;cnt++;}
  }
  if(te>=tt)v.emplace_back(tt,te);
  else vv.emplace_back(te,tt);
}
if(cnt!=0){cout << "No" << endl;return 0;}
sort(v.begin(),v.end());
sort(vv.begin(),vv.end());
reverse(vv.begin(),vv.end());
ll nw=0;
for(auto r:v){
  nw-=r.first;
  if(nw<0){cout << "No" << endl;return 0;}
  nw+=r.second;
}
for(auto r:vv){
  nw-=r.second;
  if(nw<0){cout << "No" << endl;return 0;}
  nw+=r.first;
}
cout << "Yes" << endl;
return 0;
}
