#include <bits/stdc++.h>
using namespace std;
int N, sum[20], dp[1 << 20][2], a[1 << 20];
map<int, vector<int>> pref[20];
map<int, int> stop[20];
int main() {
  cin >> N;
  for (int i = 0; i < N; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < s.size(); ++j) {
      pref[i][sum[i] += s[j] == '(' ? 1 : -1].push_back(j);
      if (!stop[i].count(sum[i])) stop[i][sum[i]] = j;
    }
    for (int j = stop[i].begin()->first + 1; j <= prev(stop[i].end())->first;
         ++j) {
      stop[i][j] = min(stop[i].count(j) ? stop[i][j] : N, stop[i][j - 1]);
    }
  }
  for (int i = 0; i < (1 << N); ++i) dp[i][1] = -1e9;
  dp[0][1] = 0;
  int ans = 0;
  for (int i = 1; i < (1 << N); ++i) {
    for (int j = 0; j < N; ++j) {
      if (~i & (1 << j)) continue;
      int b = pref[j].begin()->first + a[i ^ (1 << j)] >= 0;
      a[i] = sum[j] + a[i ^ (1 << j)];
      int s = a[i ^ (1 << j)];
      dp[i][b] = max(
          dp[i][b],
          dp[i ^ (1 << j)][1] +
              (b ? (int)(pref[j].count(-s) ? pref[j][-s].size() : 0)
                 : (int)(pref[j].count(-s)
                             ? lower_bound(pref[j][-s].begin(),
                                           pref[j][-s].end(), stop[j][-s - 1]) -
                                   pref[j][-s].begin()
                             : 0)));
    }
    ans = max(ans, dp[i][0] = max(dp[i][0], dp[i][1]));
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int N, sum[20], dp[1 << 20][2], a[1 << 20];
map<int, vector<int>> pref[20];
map<int, int> stop[20];
int main() {
  cin >> N;
  for (int i = 0; i < N; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < s.size(); ++j) {
      pref[i][sum[i] += s[j] == '(' ? 1 : -1].push_back(j);
      if (!stop[i].count(sum[i])) stop[i][sum[i]] = j;
    }
    for (int j = stop[i].begin()->first + 1; j <= prev(stop[i].end())->first;
         ++j) {
      stop[i][j] = min(stop[i].count(j) ? stop[i][j] : N, stop[i][j - 1]);
    }
  }
  for (int i = 0; i < (1 << N); ++i) dp[i][1] = -1e9;
  dp[0][1] = 0;
  int ans = 0;
  for (int i = 1; i < (1 << N); ++i) {
    for (int j = 0; j < N; ++j) {
      if (~i & (1 << j)) continue;
      int b = pref[j].begin()->first + a[i ^ (1 << j)] >= 0;
      a[i] = sum[j] + a[i ^ (1 << j)];
      int s = a[i ^ (1 << j)];
      dp[i][b] = max(
          dp[i][b],
          dp[i ^ (1 << j)][1] +
              (b ? (int)(pref[j].count(-s) ? pref[j][-s].size() : 0)
                 : (int)(pref[j].count(-s)
                             ? lower_bound(pref[j][-s].begin(),
                                           pref[j][-s].end(), stop[j][-s - 1]) -
                                   pref[j][-s].begin()
                             : 0)));
    }
    ans = max(ans, dp[i][0] = max(dp[i][0], dp[i][1]));
  }
  cout << ans << endl;
  return 0;
}

```