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>
#pragma optimize("ofast")
#pragma optimize("unroll-loops")
#pragma optimize("o3")
using namespace std;
const int N = 2e6, INF = 1e9 + 2, MOD = 1e9 + 7;
string mas[22];
int mn[22];
int en[22];
map<int, int> cnt[22];
pair<int, int> dp[N];
pair<int, int> add(int i, pair<int, int> res) {
  for (int j = 0; j < mas[i].size(); j++) {
    if (mas[i][j] == ')')
      res.first--;
    else
      res.first++;
    if (res.first < 0) {
      return res;
    }
    if (res.first == 0) res.second++;
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> mas[i];
  for (int i = 0; i < n; i++) {
    int c = 0;
    for (int j = 0; j < mas[i].size(); j++) {
      if (mas[i][j] == '(')
        c++;
      else
        c--;
      if (c < mn[i]) {
        mn[i] = c;
      }
      if (c == mn[i]) cnt[i][c]++;
    }
    en[i] = c;
  }
  for (int i = 0; i < N; i++) {
    dp[i] = {-1, -1};
  }
  dp[0] = {0, 0};
  int ans = 0;
  for (int mask = 0; mask < (1 << n); mask++) {
    if (dp[mask].first < 0) continue;
    for (int j = 0; j < n; j++) {
      if ((mask & (1 << j)) == 0) {
        int nextmask = mask ^ (1 << j);
        if (dp[mask].first >= -mn[j]) {
          pair<int, int> a = {dp[mask].first + en[j], dp[mask].second};
          if (dp[mask].first == -mn[j]) a.second += cnt[j][-dp[mask].first];
          ans = max(ans, a.second);
          if (a.first >= 0 && a.second > dp[nextmask].second) {
            dp[nextmask] = a;
          }
        } else {
          pair<int, int> a = {-1, dp[mask].second + cnt[j][-dp[mask].first]};
          ans = max(ans, a.second);
        }
      }
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
}
