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 long long N = 22;
  const long long oo = 1e18 + 7, mod = 1e9 + 7;
  string s[N];
  long long n, k;
  long long num[N][400005];
  long long ans[(1LL << N)][2];
  long long pref[(1LL << N)], mn[N];
  set<pair<long long, long long> > second;
  set<long long> nums[400005];
  void process() {
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      cin >> s[i];
      s[i] = '*' + s[i];
      second.clear();
      for (long long j = 0; j <= 400000; j++) nums[j].clear();
      long long prefix = 0;
      mn[i] = 0;
      for (long long j = 1; j < s[i].length(); j++) {
        prefix += (s[i][j] == '(' ? 1 : -1);
        mn[i] = min(mn[i], prefix);
        if (prefix <= 0) {
          second.insert({j, -prefix});
          nums[-prefix].insert(j);
        }
      }
      pref[(1LL << (i - 1))] = prefix;
      for (long long j = 400000; j >= 0; j--) {
        if (!nums[j].size()) continue;
        num[i][j] = nums[j].size();
        long long temp = (*nums[j].begin());
        set<pair<long long, long long> >::iterator it = second.find({temp, j}),
                                                   it2 = it;
        for (; it2 != second.end(); it2++)
          nums[(*it2).second].erase((*it2).first);
        second.erase(it, second.end());
      }
    }
    ans[0][1] = 0, ans[0][0] = -oo;
    for (long long i = 1; i < (1LL << n); i++) {
      ans[i][0] = ans[i][1] = -oo;
      if (__builtin_popcountll(i) >= 2) {
        for (long long j = 0; j < n; j++) {
          if (i & (1LL << j)) {
            pref[i] = (pref[i - (1LL << j)] + pref[(1LL << j)]);
            break;
          }
        }
      }
    }
    for (long long i = 1; i < (1LL << n); i++) {
      ans[i][0] = ans[i][1] = -oo;
      for (long long j = 0; j < n; j++) {
        if (!(i & (1LL << j))) continue;
        if (pref[i - (1LL << j)] + mn[j + 1] >= 0) {
          ans[i][1] = max(ans[i][1], ans[i - (1LL << j)][1] +
                                         num[j + 1][pref[i - (1LL << j)]]);
        } else
          ans[i][0] = max(ans[i][0], ans[i - (1LL << j)][1] +
                                         num[j + 1][pref[i - (1LL << j)]]);
        ans[i][0] = max(ans[i][0], ans[i - (1LL << j)][0]);
      }
    }
    cout << max(ans[(1LL << n) - 1][0], ans[(1LL << n) - 1][1]);
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    process();
  }
  