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++)
using ll = long long;
using P = pair<int,int>;

int main(){
  int n;
  cin >> n;
  string s[n];
  rep(i,n) cin >> s[i];
  vector<int> a(n), m(n);
  rep(i,n) {
    for(auto c : s[i]) {
      if(c == '(') a[i]++;
      else {
        a[i]--;
        m[i] = min(a[i],m[i]);
      } 
    }
  }
  int dif = 0;
  rep(i,n) dif += a[i];
  if(dif) {
    cout << "No" << endl;
    return 0;
  }
  int cur = 0;
  vector<P> mp, mm;
  rep(i,n) {
    if(m[i] >= 0) cur += a[i];
    else {
      if(a[i] >= 0) mp.push_back(P(m[i],a[i]));
      else if(a[i] != m[i]) mm.push_back(P(m[i],a[i]));
    }
  }
  sort(mp.begin(),mp.end(),greater<P>());
  sort(mm.begin(),mm.end());
  for(auto p : mp) {
    if(cur + p.first < 0) {
      cout << "No" << endl;
      return 0;
    }
    cur += p.second;
  } 
  for(auto p : mm) {
    if(cur + p.first < 0) {
      cout << "No" << endl;
      return 0;
    }
    cur += p.second;
  }
  cout << "Yes" << endl;
  return 0;
}
