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;
string s;
int lg[20];
int sum[20], minsum[20][400001];
vector<int> freq[20][800001];
int dp[1 << 20];
int binspref(int ind, int val) {
  int st = 0, dr = lg[ind] - 1, mij;
  while (st <= dr) {
    mij = (st + dr) >> 1;
    if (minsum[ind][mij] <= val)
      dr = mij - 1;
    else
      st = mij + 1;
  }
  return st;
}
int binsfreq(int ind, int nr, int val) {
  int st = 0, dr = freq[ind][nr].size() - 1, mij;
  while (st <= dr) {
    mij = (st + dr) >> 1;
    if (freq[ind][nr][mij] <= val)
      st = mij + 1;
    else
      dr = mij - 1;
  }
  return dr + 1;
}
int main() {
  int n, i, j, x, ans, p;
  const int c = 400000;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> s;
    lg[i] = s.size();
    x = 0;
    for (j = 0; j < s.size(); j++) {
      if (s[j] == '(')
        x++;
      else
        x--;
      freq[i][x + c].push_back(j);
      if (j > 0)
        minsum[i][j] = min(minsum[i][j - 1], x);
      else
        minsum[i][j] = x;
    }
    sum[i] = x;
  }
  ans = 0;
  for (i = 1; i < (1 << n); i++) {
    x = 0;
    for (j = 0; j < n; j++)
      if (i & (1 << j)) x = x + sum[j];
    for (j = 0; j < n; j++)
      if (i & (1 << j)) {
        if (minsum[j][lg[j] - 1] + x - sum[j] >= 0)
          dp[i] = max(
              dp[i], dp[i ^ (1 << j)] + int(freq[j][-(x - sum[j]) + c].size()));
        else {
          p = binspref(j, -x + sum[j] - 1);
          p--;
          ans = max(ans, dp[i ^ (1 << j)] + binsfreq(j, -(x - sum[j]) + c, p));
        }
      }
    ans = max(ans, dp[i]);
  }
  cout << ans;
  return 0;
}
