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>
const int N = 400005;
const int M = 21;
std::vector<int> val[M], cnt[M], mn[M], pos[M];
int cnt1[1 << (M - 1)], f[1 << (M - 1)][2];
bool valid[1 << (M - 1)];
inline int lb(int x) { return x & (-x); }
void init(int x) {
  static char s[N];
  static int pre[N << 1];
  scanf("%s", s + 1);
  int sz = strlen(s + 1), sum = 0;
  for (int i = 1; i <= sz; i++) {
    if (s[i] == '(')
      sum++;
    else
      sum--;
    val[x].push_back(sum);
    if (!mn[x].size())
      mn[x].push_back(sum);
    else
      mn[x].push_back(std::min(mn[x].back(), sum));
    if (mn[x].back() == val[x].back())
      pos[x].push_back(i - 1);
    else
      pos[x].push_back(pos[x].back());
  }
  for (int i = 0; i < val[x].size(); i++) {
    cnt[x].push_back(++pre[val[x][i] + N - 5]);
  }
  for (int i = 0; i < val[x].size(); i++) pre[val[x][i] + N - 5] = 0;
}
int calc(int x, int k) {
  int l = 0, r = mn[x].size() - 1, ans = 0;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (mn[x][mid] >= k)
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return mn[x][ans] == k ? cnt[x][pos[x][ans]] : 0;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) init(i);
  for (int i = 1; i <= n; i++) cnt1[1 << (i - 1)] = val[i].back();
  for (int i = 1; i < (1 << n); i++) cnt1[i] = cnt1[i ^ lb(i)] + cnt1[lb(i)];
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j < n; j++) {
      if (!((i >> j) & 1)) {
        f[i ^ (1 << j)][1] = std::max(f[i][1], f[i ^ (1 << j)][1]);
        if (valid[i] || i == 0) {
          if (cnt1[i] + mn[j + 1].back() >= 0)
            f[i ^ (1 << j)][0] =
                std::max(f[i ^ (1 << j)][0], f[i][0] + calc(j + 1, -cnt1[i])),
                        valid[i ^ (1 << j)] = 1;
          else
            f[i ^ (1 << j)][1] =
                std::max(f[i ^ (1 << j)][1], f[i][0] + calc(j + 1, -cnt1[i]));
        }
      }
    }
  }
  printf("%d\n", std::max(f[(1 << n) - 1][0], f[(1 << n) - 1][1]));
}
