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;
  namespace Ruri {
  const int infi = 1e9;
  const long long infl = 4e18;
  inline long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
  inline long long pos_mod(long long a, long long b) { return (a % b + b) % b; }
  }  // namespace Ruri
  using namespace Ruri;
  namespace Read {
  inline int ri() {
    int x;
    scanf("%d", &x);
    return x;
  }
  inline long long rl() {
    long long x;
    scanf("%lld", &x);
    return x;
  }
  inline double rd() {
    double x;
    scanf("%lf", &x);
    return x;
  }
  }  // namespace Read
  namespace DeBug {}
  using namespace DeBug;
  int main() {
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<string> s(n);
    vector<int> len(n), delta(n), mn_delta(n, 0);
    vector<vector<vector<int> > > pos(n);
    for (int i = 0, bbb = n - 1; i <= bbb; ++i) {
      cin >> s[i];
      len[i] = s[i].length();
      pos[i].resize(2 * len[i] + 1);
      int now = len[i];
      for (int j = 0, bbb = len[i] - 1; j <= bbb; ++j) {
        now += (s[i][j] == '(' ? 1 : -1);
        pos[i][now].push_back(j + 1),
            mn_delta[i] = min(mn_delta[i], now - len[i]);
      }
      delta[i] = now - len[i];
    }
    int ans = 0;
    vector<int> dp(1 << n, -1);
    dp[0] = 0;
    for (int mask = 0, bbb = (1 << n) - 1; mask <= bbb; ++mask)
      if (~dp[mask]) {
        int psum = 0;
        for (int bit = 0, bbb = n - 1; bit <= bbb; ++bit)
          if ((mask >> bit) & 1) psum += delta[bit];
        for (int bit = 0, bbb = (n - 1); bit <= bbb; ++bit)
          if (!((mask >> bit) & 1)) {
            int balance_pos = len[bit] - psum;
            if (psum + mn_delta[bit] < 0) {
              int bound = pos[bit][balance_pos - 1][0];
              int res = (int)(lower_bound(pos[bit][balance_pos].begin(),
                                          pos[bit][balance_pos].end(), bound) -
                              pos[bit][balance_pos].begin());
              ans = max(ans, dp[mask] + res);
              continue;
            }
            int add = 0;
            if (0 <= balance_pos && balance_pos <= 2 * len[bit])
              add = (int)pos[bit][balance_pos].size();
            dp[mask | (1 << bit)] = max(dp[mask | (1 << bit)], dp[mask] + add);
          }
      }
    printf("%d\n", max(ans, dp.back()));
    return 0;
  }
  