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>
  template <class C>
  C& mini(C& a, C b) {
    if (b < a) a = b;
    return a;
  }
  template <class C>
  C& maxi(C& a, C b) {
    if (a < b) a = b;
    return a;
  }
  template <typename T>
  void read(std::vector<T>& A) {
    for (T& x : A) std::cin >> x;
  }
  template <typename T>
  void readv(std::vector<T>& A) {
    int N;
    std::cin >> N;
    A.resize(N);
    read(A);
  }
  using namespace std;
  struct Bracket {
    string S;
    int tot;
    int low;
    vector<int> cnt;
    void init() {
      int c = 0;
      low = 0;
      cnt.resize(1);
      for (char x : S) {
        if (x == '(') {
          c++;
        } else {
          c--;
          if (c <= low) {
            if (c < low) {
              low = c;
              cnt.resize(-c + 1);
            }
            cnt[-c]++;
          }
        }
      }
      tot = c;
    }
    int solve(int lvl) {
      if (lvl < cnt.size()) return cnt[lvl];
      return 0;
    }
  };
  int N;
  Bracket brackets[20];
  int sumtot[1 << 20];
  int dp[1 << 20];
  int solve(int mask) {
    int& ret = dp[mask];
    if (ret >= 0) return ret;
    int lvl = sumtot[((1 << N) - 1) ^ mask];
    if (lvl < 0) return ret = 0;
    ret = 0;
    for (int i = 0; i < N; i++)
      if (mask & (1 << i)) {
        int r = brackets[i].solve(lvl);
        if (lvl + brackets[i].low >= 0) r += solve(mask ^ (1 << i));
        maxi(ret, r);
      }
    return ret;
  }
  int main(int argc, const char* argv[]) {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.precision(20);
    cout.setf(ios::fixed);
    cin >> N;
    for (int i = 0; i < N; i++) {
      auto& b = brackets[i];
      cin >> b.S;
      b.init();
    }
    for (int m = 1; m < (1 << N); m++) {
      int lb = __builtin_ctz(m);
      sumtot[m] = sumtot[m ^ (1 << lb)] + brackets[lb].tot;
    }
    memset(dp, -1, sizeof(dp[0]) << N);
    cout << solve((1 << N) - 1) << endl;
    return 0;
  }
  