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>
  const int N = 500005, inf = 0x3f3f3f3f;
  using namespace std;
  int n, f[1 << 20], sm[1 << 20], a[25][N], b[25][N], c[25][N], ans, len[25];
  char s[25][N];
  void init() {
    for (register int i = 1; i <= n; i++) {
      for (register int j = 1; j <= len[i]; j++) {
        a[i][j] = a[i][j - 1] + (s[i][j] == '(' ? 1 : -1);
        b[i][j] = b[i][j - 1], c[i][j] = c[i][j - 1];
        if (a[i][j] < b[i][j])
          b[i][j] = a[i][j], c[i][j] = 1;
        else if (a[i][j] == b[i][j])
          c[i][j]++;
      }
    }
  }
  int get(int i, int mn) {
    int l = 0, r = len[i], res = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (b[i][mid] >= mn)
        res = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    return res;
  }
  void solve() {
    memset(f, -inf, sizeof f), f[0] = 0;
    for (register int s = 1; s <= (1 << n) - 1; s++) {
      for (register int i = 1; i <= n; i++) {
        if (!((s >> (i - 1)) & 1)) continue;
        int t = s ^ (1 << (i - 1));
        if (f[t] == -inf) continue;
        int mn = sm[t] + b[i][len[i]];
        if (mn >= 0) {
          if (f[s] < f[t] + (mn ? 0 : c[i][len[i]])) {
            f[s] = f[t] + (mn ? 0 : c[i][len[i]]);
            sm[s] = sm[t] + a[i][len[i]];
          }
        }
        mn = -sm[t];
        int p = get(i, mn);
        ans = max(ans, f[t] + (b[i][p] == mn ? c[i][p] : 0));
      }
    }
  }
  signed main() {
    scanf("%d", &n);
    for (register int i = 1; i <= n; i++) {
      scanf("%s", s[i] + 1), len[i] = strlen(s[i] + 1);
    }
    init();
    solve();
    printf("%d\n", ans);
    return 0;
  }
  