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;
#define REP(i, n) for(int i=0; i<(int)(n); i++)
typedef pair<int, int> PAIR;
int main(){
int N; cin >> N; string S[N]; vector<PAIR> U, V;
REP(i, N){ 
  cin >> S[i];
  int l=0, m=0, n=S[i].size();
  REP(j, n){ l=(S[i][j]=='(')?l+1:l-1; m=min(m, l); }
  if(l>=0){ U.push_back(make_pair(-m, l)); }
  else{ V.push_back(make_pair(-(m-l), -l)); }
}
sort(U.begin(), U.end()); sort(V.begin(), V.end());
int l=0, m=0, u=U.size(), v=V.size();
REP(i, u){ if(l-U[i].first>=0){ l=l+U[i].second; }else{ cout << "No\n"; return 0; } }
REP(i, v){ if(m-V[i].first>=0){ m=m+V[i].second; }else{ cout << "No\n"; return 0; } }
if(l==m){ cout << "Yes\n"; }else{ cout << "No\n"; }
}