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: import sys
  
  inp = sys.stdin.readline
  
  def solve():
      n = int(inp())
      a = [[0] for i in range(n)]
      mb = [0] * (1 << n)
      for i in range(n):
          s = inp().strip()
          m = 0
          b = 0
          for c in s:
              if c == '(':
                  b += 1
              else:
                  b -= 1
                  if b <= 0:
                      if m > b:
                          m = b
                          a[i].append(1)
                      elif m == b:
                          a[i][-1] += 1
          mb[1 << i] = b
      dp = [None] * (1 << n)
      dp[0] = 0
      ans = 0
      for i in range(1, 1 << n):
          j = i & (i - 1)
          if j:
              mb[i] = mb[i-j] + mb[j]
          nd = None
          for j in range(n):
              if (i >> j) & 1:
                  d = dp[i - (1 << j)]
                  if d is not None:
                      bb = mb[i - (1 << j)]
                      m = len(a[j]) - 1
                      if bb - m < 0:
                          ans = max(ans, a[j][bb] + d)
                      elif bb - m == 0:
                          if nd is None:
                              nd = a[j][-1] + d
                          else:
                              nd = max(nd, a[j][-1] + d)
                      else:
                          if nd is None:
                              nd = d
                          else:
                              nd = max(nd, d)
          if nd is not None:
              ans = max(ans, nd)
          dp[i] = nd
      print(ans)
  
  def main():
      solve()
  
  if __name__ == '__main__':
      main()
  