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 INF = 0x3f3f3f3f;
  const int64_t INFLL = 0x3f3f3f3f3f3f3f3f;
  const int64_t MOD = 1e9 + 7;
  const double EPS = 1e-9;
  const double PI = acos(-1);
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int64_t rand(int64_t x, int64_t y) {
    return uniform_int_distribution<int64_t>(x, y)(rng);
  }
  int64_t mul(int64_t a, int64_t b) {
    int64_t ret = (1ll * (a % MOD) * (b % MOD)) % MOD;
    return (ret + MOD) % MOD;
  }
  int64_t add(int64_t a, int64_t b) {
    int64_t ret = (1ll * (a % MOD) + (b % MOD)) % MOD;
    return (ret + MOD) % MOD;
  }
  int64_t pow_exp(int64_t n, int64_t p) {
    if (!p) return 1;
    if (p & 1) return mul(n, pow_exp(n, p - 1));
    int64_t tmp = pow_exp(n, p / 2);
    return mul(tmp, tmp);
  }
  const int64_t MAX = 2e5 + 10;
  const int64_t NMAX = 2e5 + 10;
  const int64_t MMAX = 2e5 + 10;
  const int64_t LOG_MAX = ceil(log2(double(NMAX)));
  const int64_t BLOCK = ceil(sqrt(double(NMAX)));
  int64_t n;
  string str[20];
  map<int64_t, vector<int64_t> > mp[20];
  int64_t tot[20];
  int64_t mem[1 << 20];
  int64_t go(int64_t msk, int64_t sm) {
    if (msk + 1 == 1 << n) {
      return 0;
    }
    int64_t &ret = mem[msk];
    if (ret + 1) return ret;
    int64_t ans = 0;
    for (int64_t i = 0; i < n; i++)
      if (((msk >> i) & 1) == 0) {
        int64_t mn = mp[i].begin()->first;
        if (sm + mn >= 0) {
          int64_t cans = 0;
          cans += go(msk ^ (1 << i), sm + tot[i]);
          auto it = mp[i].find(-sm);
          if (it != mp[i].end()) cans += it->second.size();
          ans = max(ans, cans);
        } else {
          auto it = mp[i].find(-sm);
          if (it == mp[i].end()) {
            continue;
          }
          if (it == mp[i].begin()) {
            int64_t cans = it->second.size();
            ans = max(ans, cans);
          } else {
            auto &v = mp[i][-sm];
            int64_t r = (--it)->second.front();
            int64_t cans = lower_bound(v.begin(), v.end(), r) - v.begin();
            ans = max(ans, cans);
          }
        }
      }
    return ret = ans;
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    cin >> n;
    for (int64_t i = 0; i < n; i++) {
      cin >> str[i];
    }
    for (int64_t i = 0; i < n; i++) {
      int64_t sm = 0;
      for (int64_t j = 0; j < str[i].size(); j++) {
        sm += str[i][j] == '(' ? +1 : -1;
        mp[i][sm].push_back(j);
      }
      tot[i] = sm;
    }
    memset(mem, -1, sizeof mem);
    cout << go(0, 0) << '\n';
  }
  