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>
  constexpr int maxn = 20, maxm = 4e5 + 10, inf = 0x3f3f3f3f;
  char str[maxn][maxm];
  int n, cnt[maxn][maxm], L[maxn], R[maxn];
  void init(const int id) {
    std::stack<char> stk;
    const auto len = (int)std::strlen(str[id] + 1);
    for (auto lim_i = len, i = 1; i <= lim_i; i++) {
      const auto c = str[id][i];
      if (c == '(')
        stk.push(c);
      else if (c == ')') {
        if (!stk.empty() && stk.top() == '(')
          stk.pop();
        else
          stk.push(c);
      }
      if (stk.empty() || stk.top() == ')') cnt[id][stk.size()]++;
    }
    while (!stk.empty()) {
      const auto c = stk.top();
      stk.pop();
      if (c == ')')
        L[id]++;
      else
        R[id]++;
    }
  }
  struct status {
    int rem = -inf, v;
  } f[1 << maxn];
  bool vis[1 << maxn];
  void solve() {
    std::cin >> n;
    for (auto lim_i = n - 1, i = 0; i <= lim_i; i++)
      std::cin >> (str[i] + 1), init(i);
    for (auto lim_i = n - 1, i = 0; i <= lim_i; i++)
      std::cerr << L[i] << " " << R[i] << "\n";
    std::queue<int> q;
    q.push(0);
    f[0].rem = 0;
    int ans = 0;
    while (!q.empty()) {
      const auto s = q.front();
      q.pop();
      ans = std::max(ans, f[s].v);
      for (auto lim_nxt = n - 1, nxt = 0; nxt <= lim_nxt; nxt++)
        if (!((s >> nxt) & 1)) {
          if (f[s].rem < L[nxt]) {
            const auto t = f[s].v + cnt[nxt][f[s].rem];
            ans = std::max(ans, t);
            continue;
          }
          const auto p = s | (1 << nxt);
          const auto r = f[s].rem - L[nxt] + R[nxt];
          f[p].rem = r;
          const auto t = f[s].v + cnt[nxt][f[s].rem];
          if (t > f[p].v) f[p].v = t;
          if (!vis[p]) {
            q.push(p);
            vis[p] = 1;
          }
        }
    }
    std::cout << ans << "\n";
  }
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int t = 1;
    for (auto REPN_lIM = t, REPN = 1; REPN <= REPN_lIM; REPN++) solve();
    return (0 - 0);
  }
  