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.
1
)(((((()((())(((((((
 1
)()())((()))((()((()
 10
((((((((((((((((()))))))))))))))))
(((((((((((((((())))))))))))))))
((((((((((((((((((((()))))))))))))))))))))
(((((((((((((((((((((())))))))))))))))))))))
()()()()()()()()()()()()()()()()()()()()
()()()()()()()()()()()()()()()()()()()()()()()()()()
(((((((((((((((((())))))))))))))))))
()()()()()()()()()()()()()()()()()()()
()()()()()()()()()()()()()()()()()()
()()()()()()()()()()()()()()()()()()()()()()()
 20
)
(
)
)
)
)
)
(
(
)
)
)
)
(
)
)
)
)
(
)
 6
()()()()()
()()()()
()
((((()))))
((()))
()()
 1
((((((((((((((((((((
 10
((((((((((((((((((((((((()))))))))))))))))))))))))
()()()()()()()()()()()()()()()()()()()
(((((((((((((((((())))))))))))))))))
()()()()()()()()()()()()()()()()()()()()()()
((((((((((((((((((()))))))))))))))))))
((((((((((((((((((((((()))))))))))))))))))))))
()()()()()()()()()()()()()()()()()()()
((((((((((((((()))))))))))))))
()()()()()()()()()()()()()()()()()()()()()()()()
()()()()()()()()()()()()()()()()
 2
))())))((()
()((((((()))))))(()
 20
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
 20
(
(
)
(
(
)
(
(
(
(
(
(
)
)
(
(
(
(
(
)
 10
()()
(())
()()
()()
(())
(((())))
()
(())
()()
()
 1
))))))))))))))))))))
 6
(((())))
(((())))
()()()
(())
()()()()()
()()
 10
)
)
))
))
))
))
))
))
)))
)))
 10
)()
()
)
)
)
)(
)))
)(
(((
))
 20
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
 10
))
()(
(
()
()
((((
(
(
(((
(
 10
()
()()()()
(())
((()))
(())
()()
(())
()()
()
()
 10
())
()
)
)))
))
()
)
))
)))
)
 1
))))))()))())(()))))
 10
((
(((
(
(((
((((
((
(
(
(
((
 20
)
(
(
)
)
(
)
(
(
(
)
)
)
)
(
)
(
(
)
(

0
 0
 111
 5
 14
 0
 105
 3
 0
 5
 14
 0
 13
 0
 5
 0
 4
 15
 3
 0
 0
 10

#include <bits/stdc++.h>
using namespace std;
void solve_problem() {
  const int offset = 4 * 1e5 + 10;
  int n;
  cin >> n;
  vector<vector<int>> b(n), lvp(n), m(n);
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    int prev_b = 0;
    b[i].resize(s.size(), 0);
    lvp[i].resize(1e6, 0);
    m[i].resize(s.size(), 0);
    m[i][0] = 2;
    for (int j = 0; j < s.size(); j++) {
      if (s[j] == '(') {
        b[i][j] = prev_b + 1;
      } else {
        b[i][j] = prev_b - 1;
      }
      prev_b = b[i][j];
      m[i][j] = min(m[i][j - 1], b[i][j]);
      if (m[i][j] == b[i][j]) {
        ++lvp[i][b[i][j] + offset];
      }
    }
  }
  vector<pair<array<int, 2>, int>> dp(
      (1 << n), make_pair(array<int, 2>{int(-1e9), int(-1e9)}, 0));
  int max_num = 0;
  for (int mask = 1; mask < (1 << n); mask++) {
    for (int j = 0; j < n; j++) {
      if (mask & (1 << j)) {
        int sub_mask = mask ^ (1 << j);
        int cur_balance = dp[sub_mask].second;
        int cur_num = lvp[j][offset - cur_balance];
        int size = m[j].size() - 1;
        dp[mask].second = cur_balance + b[j][size];
        dp[mask].first[0] = max(dp[mask].first[0], dp[sub_mask].first[0]);
        if (cur_balance + m[j][size] < 0) {
          if (sub_mask != 0) {
            dp[mask].first[0] =
                max(dp[mask].first[0], dp[sub_mask].first[1] + cur_num);
          } else {
            dp[mask].first[0] = max(dp[mask].first[0], cur_num);
          }
        } else {
          if (sub_mask != 0) {
            dp[mask].first[1] =
                max(dp[mask].first[1], dp[sub_mask].first[1] + cur_num);
          } else {
            dp[mask].first[1] = max(dp[mask].first[1], cur_num);
          }
        }
        max_num = max(max_num, dp[mask].first[0]);
        max_num = max(max_num, dp[mask].first[1]);
      }
    }
  }
  cout << max(max_num, 0) << endl;
}
int main() {
  int t = 1;
  for (int i = 0; i < t; i++) {
    solve_problem();
  }
  return 0;
}
