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;
const int maxn = 21;
int n, m, ans, tot;
int f[1 << maxn], mn[maxn], cnt[maxn], sum[1 << maxn], tmp[maxn];
vector<int> w[maxn];
string s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int now = 0;
    cin >> s, mn[i] = 0, cnt[i] = 0;
    for (int j = 0; j < s.size(); j++) {
      if (s[j] == '(')
        now++;
      else {
        now--;
        if (now < mn[i])
          mn[i] = now, cnt[i] = 1;
        else if (now == mn[i])
          cnt[i]++;
      }
      tmp[j] = now;
    }
    sum[1 << (i - 1)] = now;
    w[i].resize(800000);
    int tmpmn = 0;
    for (int j = 0; j < s.size(); j++) {
      tmpmn = min(tmpmn, tmp[j]);
      if (tmpmn == tmp[j]) w[i][400000 + tmp[j]]++;
    }
  }
  f[0] = 0;
  for (int i = 1; i < (1 << n); i++) {
    sum[i] = sum[i ^ i & -i] + sum[i & -i];
    for (int j = 1; j <= n; j++)
      if ((i >> (j - 1)) & 1) {
        int pre = (i ^ (1 << (j - 1))), v = f[pre];
        if (sum[pre] < 0) continue;
        if (sum[pre] == -mn[j])
          v += cnt[j];
        else if (sum[pre] < -mn[j])
          v = -1;
        f[i] = max(f[i], v);
        if (sum[pre] >= 0) ans = max(ans, f[pre] + w[j][400000 - sum[pre]]);
      }
    ans = max(ans, f[i]);
  }
  printf("%d\n", ans);
  return 0;
}
