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;
  const int64_t mod = 1e9 + 7;
  void solve() {
    int64_t n;
    cin >> n;
    vector<string> s(n);
    for (int64_t i = 0; i < n; i++) cin >> s[i];
    vector<vector<int64_t>> mp(n), mp_cnt(n);
    vector<int64_t> str_bal(n);
    for (int64_t i = 0; i < n; i++) {
      mp[i].resize(s[i].size());
      mp_cnt[i].resize(s[i].size());
      int64_t bal = 0, Min = 1;
      for (int64_t j = 0; j < s[i].size(); j++) {
        if (s[i][j] == '(')
          bal++;
        else
          bal--;
        Min = min(Min, bal);
        mp[i][j] = Min;
        if (j > 0 && mp[i][j - 1] == Min) {
          if (bal == Min)
            mp_cnt[i][j] = mp_cnt[i][j - 1] + 1;
          else
            mp_cnt[i][j] = mp_cnt[i][j - 1];
        } else
          mp_cnt[i][j] = 1;
      }
      str_bal[i] = bal;
    }
    int64_t msk = 1 << n;
    vector<int64_t> dp(msk, -1), msk_bal(msk);
    dp[0] = 0;
    int64_t ans = 0;
    for (int64_t i = 1; i < msk; i++) {
      for (int64_t j = 0; j < n; j++) {
        int64_t to = i ^ (1 << j);
        if (to == i || dp[to] == -1) continue;
        msk_bal[i] = msk_bal[to] + str_bal[j];
        int64_t l = 0, r = s[j].size();
        while (l < r) {
          int64_t mid = (l + r) / 2;
          if (mp[j][mid] >= -msk_bal[to])
            l = mid + 1;
          else
            r = mid;
        }
        if (l == s[j].size()) {
          if (mp[j].back() == -msk_bal[to])
            dp[i] = max(dp[i], dp[to] + mp_cnt[j].back());
          else
            dp[i] = max(dp[i], dp[to]);
        } else
          ans = max(ans, dp[to] + (l > 0 ? mp_cnt[j][l - 1] : 0));
      }
    }
    ans = max(ans, dp.back());
    cout << ans << "\n";
    return;
  }
  signed main() {
    int64_t t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  