description: A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence (or, shortly, an RBS) is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example:
  
    * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"); 
    * bracket sequences ")(", "(" and ")" are not. 
  
  
  
  Let's denote the concatenation of two strings x and y as x+y. For example, "()()" + ")(" = "()())(".
  
  You are given n bracket sequences s_1, s_2, ..., s_n. You can rearrange them in any order (you can rearrange only the strings themselves, but not the characters in them).
  
  Your task is to rearrange the strings in such a way that the string s_1 + s_2 + ... + s_n has as many non-empty prefixes that are RBS as possible.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 20).
  
  Then n lines follow, the i-th of them contains s_i — a bracket sequence (a string consisting of characters "(" and/or ")". All sequences s_i are non-empty, their total length does not exceed 4 ⋅ 10^5.
  
  Output
  
  Print one integer — the maximum number of non-empty prefixes that are RBS for the string s_1 + s_2 + ... + s_n, if the strings s_1, s_2, ..., s_n can be rearranged arbitrarily.
  
  Examples
  
  Input
  
  
  2
  (
  )
  
  
  Output
  
  
  1
  
  
  Input
  
  
  4
  ()()())
  (
  (
  )
  
  
  Output
  
  
  4
  
  
  Input
  
  
  1
  (())
  
  
  Output
  
  
  1
  
  
  Input
  
  
  1
  )(()
  
  
  Output
  
  
  0
  
  Note
  
  In the first example, you can concatenate the strings as follows: "(" + ")" = "()", the resulting string will have one prefix, that is an RBS: "()".
  
  In the second example, you can concatenate the strings as follows: "(" + ")" + "()()())" + "(" = "()()()())(", the resulting string will have four prefixes that are RBS: "()", "()()", "()()()", "()()()()".
  
  The third and the fourth examples contain only one string each, so the order is fixed.
solution: #include <bits/stdc++.h>
  using namespace std;
  template <class _F, class _S>
  ostream &operator<<(ostream &out, const pair<_F, _S> &_pair) {
    return out << "(" << _pair.first << "," << _pair.second << ")";
  }
  template <class T>
  ostream &operator<<(ostream &out, const vector<T> &_vec) {
    out << "[";
    for (int _vecp = 0; _vecp < _vec.size(); ++_vecp)
      out << _vec[_vecp] << ",]"[_vecp + 1 == _vec.size()];
    return out;
  }
  void debug() {}
  template <class T, class... U>
  void debug(const T &head, const U &...tail) {
    cerr << head << (sizeof...(tail) ? " " : "\n"), debug(tail...);
  }
  template <class... T>
  void D(const T &...parm) {
    cerr << ">>> ", debug(parm...);
  }
  const int mx = 4e5 + 10;
  const long long mod = 998244353;
  mt19937_64 mtrand(time(0));
  int n, m, T;
  string s[20];
  int b[20];
  int cnt[1 << 20];
  vector<int> vec[20];
  unordered_map<int, int> mp[20], mp2[20];
  int dp[1 << 20][2];
  int sk[mx], tot;
  void solve(int id) {
    for (int i = 0; i < s[id].size(); ++i) {
      if (s[id][i] == '(')
        ++b[id];
      else
        --b[id];
      vec[id].push_back(b[id]);
      ++mp2[id][b[id]];
      if (!mp[id].count(b[id])) mp[id][b[id]] = mp2[id][b[id] + 1];
    }
  }
  void run_case() {
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> s[i];
      solve(i);
    }
    int up = 1 << n;
    for (int i = 0; i < up; ++i) {
      for (int j = 0; j < n; ++j) {
        if ((i >> j) & 1) cnt[i] += b[j];
      }
      dp[i][0] = dp[i][1] = -0x3f3f3f3f;
    }
    dp[0][0] = 0;
    for (int i = 1; i < up; ++i) {
      for (int j = 0; j < n; ++j) {
        if ((i >> j) & 1) {
          int st = i & (up - (1 << j) - 1);
          if (mp[j].count(-cnt[st] - 1)) {
            dp[i][1] = max(dp[i][1], dp[st][0] + mp[j][-cnt[st] - 1]);
          } else {
            dp[i][0] = max(dp[i][0], dp[st][0] + mp2[j][-cnt[st]]);
          }
          dp[i][1] = max(dp[i][1], dp[st][1]);
        }
      }
    }
    long long ans = max(dp[up - 1][0], dp[up - 1][1]);
    cout << ans << "\n";
  }
  int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    T = 1;
    for (int i = 1; i <= T; ++i) run_case();
  }
  