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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx")
  using namespace std;
  const int N = 1e6 + 10;
  const long long mod = 1e9 + 7;
  const long long inf = 8e18;
  const int LOG = 20;
  long long pw(long long a, long long b, long long M = mod, long long ret = 1) {
    while (b) {
      ret = ret * (b & 1 ? a : 1) % M, a = a * a % M, b >>= 1;
    }
    return ret;
  }
  int n, A[N], mn[1 << LOG], sum[1 << LOG], dp[2][1 << LOG];
  char C[N];
  string s[LOG];
  vector<int> cnt[N];
  int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
      scanf("%s", C);
      s[i] = C;
      cnt[i].resize((int)s[i].size() + 1, 0);
      for (int j = 0; j < (int)s[i].size(); j++) {
        A[i] += (s[i][j] == '(' ? 1 : -1);
        mn[i] = min(mn[i], A[i]);
        if (mn[i] == A[i]) cnt[i][-A[i]]++;
      }
    }
    fill(dp[0], dp[0] + (1 << LOG), -1e9);
    for (int mask = 1; mask < 1 << n; mask++) {
      sum[mask] =
          sum[mask - (1 << (__builtin_ctz(mask)))] + A[__builtin_ctz(mask)];
    }
    dp[0][0] = dp[1][0] = 0;
    int tot = 0;
    for (int mask = 1; mask < 1 << n; mask++) {
      for (int i = 0; i < n; i++) {
        if (!(mask >> i & 1)) {
          continue;
        }
        int mask2 = mask ^ (1 << i);
        if (sum[mask2] < 0) continue;
        if (sum[mask2] >= -mn[i]) {
          dp[0][mask] =
              max(dp[0][mask],
                  dp[0][mask2] +
                      ((sum[mask2] > (int)s[i].size()) ? 0 : cnt[i][sum[mask2]]));
        }
        dp[1][mask] =
            max(dp[1][mask],
                dp[0][mask2] +
                    ((sum[mask2] > (int)s[i].size()) ? 0 : cnt[i][sum[mask2]]));
      }
      tot = max(tot, dp[1][mask]);
    }
    printf("%d", tot);
    return 0;
  }
  