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 IO {
  int read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 1) + (x << 3) + (ch ^ 48);
      ch = getchar();
    }
    return x * f;
  }
  void write(int x, char sp) {
    char ch[20];
    int len = 0;
    if (x < 0) x = -x, putchar('-');
    do {
      ch[len++] = x % 10 + '0';
      x /= 10;
    } while (x);
    for (int i = len - 1; ~i; i--) putchar(ch[i]);
    putchar(sp);
  }
  void ckmin(int& x, int y) { x = x < y ? x : y; }
  void ckmax(int& x, int y) { x = x > y ? x : y; }
  }  // namespace IO
  using namespace IO;
  int n, U, ans, mn[30], val[30], sum[1 << 20], buc[800010], tim[800010][21],
      f[1 << 20];
  const int bs = 400000;
  char ch[30];
  signed main() {
    n = read();
    U = (1 << n) - 1;
    for (int i = 1; i <= n; i++) {
      scanf("%s", ch + 1);
      memset(buc, 0, sizeof(buc));
      int m = 0, now = 0, len = strlen(ch + 1);
      for (int j = 1; j <= len; j++) {
        now += ch[j] == '(' ? 1 : -1;
        if (m > now) tim[bs + m][i] = buc[bs + m], --m;
        ++buc[bs + now];
      }
      tim[bs + m][i] = buc[bs + m];
      mn[i] = m;
      val[i] = now;
    }
    for (int i = 1; i <= U; i++)
      for (int j = 0; j < n; j++)
        if (i & (1 << j)) sum[i] += val[j + 1];
    memset(f, 0xc0, sizeof(f));
    f[0] = 0;
    for (int i = 0; i <= U; i++)
      for (int j = 1; j <= n; j++)
        if (!(i & (1 << j - 1)) && sum[i] + mn[j] >= 0)
          ckmax(f[i | (1 << j - 1)], f[i] + tim[bs - sum[i]][j]);
    for (int i = 0; i < +U; i++)
      if (sum[i] >= 0) {
        ckmax(ans, f[i]);
        for (int j = 1; j <= n; j++)
          if (!(i & (1 << j - 1))) ckmax(ans, f[i] + tim[bs - sum[i]][j]);
      }
    write(ans, '\n');
    return 0;
  }
  