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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using vpii = vector<pii>;
using vint = vector<int>;
using ar3 = array<int, 3>;
using ar4 = array<int, 4>;
using va3 = vector<ar3>;
using va4 = vector<ar4>;
const int N = 4e5 + 5;
const int Inf = 1e9 + 7;
const ll inf = 1e18 + 7;
int n;
map<int, int> m[20];
vint pos[20][N * 2];
int sb[20];
inline int get_bal(int msk) {
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    if ((msk >> i) & 1) ans += sb[i];
  }
  return ans;
}
int dp[1 << 20];
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    int bal = 0;
    int idx = 0;
    for (char c : s) {
      if (c == '(')
        ++bal;
      else
        --bal;
      pos[i][bal + N].push_back(idx);
      ++idx;
    }
    sb[i] = bal;
  }
  for (int i = 0; i < 1 << n; ++i) {
    dp[i] = -Inf;
  }
  dp[0] = 0;
  for (int i = 0; i < (1 << n); ++i) {
    for (int j = 0; j < n; ++j) {
      if ((i >> j) & 1) continue;
      int tmp = -get_bal(i);
      dp[i | (1 << j)] = max(dp[i | (1 << j)], dp[i]);
      if (tmp <= 0) {
        int cnt = int((pos[j][N + tmp]).size());
        if (int((pos[j][N + tmp - 1]).size())) {
          cnt = lower_bound((pos[j][N + tmp]).begin(), (pos[j][N + tmp].end()),
                            pos[j][N + tmp - 1][0]) -
                pos[j][N + tmp].begin();
        }
        dp[i | (1 << j)] = max(dp[i | (1 << j)], dp[i] + cnt);
      }
    }
  }
  cout << dp[(1 << n) - 1] << '\n';
}
signed main() {
  int tt = 1;
  ios_base::sync_with_stdio(0);
  cout << setprecision(11) << fixed;
  cin.tie(NULL);
  while (tt--) {
    solve();
  }
}
