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 pii = pair<int, int>;
using pll = pair<ll, ll>;
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9 + 7;
const int MAXN = 8e5 + 5;
const int INF = 2e9;
vector<int> vals[21][MAXN], ss;
int dp[1 << 20][2], sm[1 << 20];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int offset = 4e5;
  int n;
  cin >> n;
  string s;
  for (int i = 0; i < n; i++) {
    cin >> s;
    int tam = s.size(), sum = 0;
    for (int j = 0; j < tam; j++) {
      if (s[j] == '(')
        sum++;
      else
        sum--;
      vals[i][sum + offset].push_back(j);
    }
    ss.push_back(sum + offset);
  }
  for (int i = 0; i < (1 << n); i++) {
    int sum = 0;
    for (int j = 0; j < n; j++) {
      if (i & (1 << j)) sum += ss[j] - offset;
    }
    sm[i] = sum + offset;
  }
  for (int i = 1; i < (1 << n); i++) {
    for (int j = 0; j < n; j++) {
      if (i & (1 << j)) {
        dp[i][1] = max(dp[i][1], dp[i ^ (1 << j)][1]);
        if (sm[i ^ (1 << j)] < offset) continue;
        if (vals[j][offset - (sm[i ^ (1 << j)] - offset) - 1].size()) {
          int np = vals[j][offset - (sm[i ^ (1 << j)] - offset) - 1][0];
          int tot =
              vals[j][offset - (sm[i ^ (1 << j)] - offset)].end() -
              lower_bound(vals[j][offset - (sm[i ^ (1 << j)] - offset)].begin(),
                          vals[j][offset - (sm[i ^ (1 << j)] - offset)].end(),
                          np);
          tot = vals[j][offset - (sm[i ^ (1 << j)] - offset)].size() - tot;
          dp[i][1] = max(dp[i][1], dp[i ^ (1 << j)][0] + tot);
        } else {
          dp[i][0] = max(
              dp[i][0],
              dp[i ^ (1 << j)][0] +
                  (int)vals[j][offset - (sm[i ^ (1 << j)] - offset)].size());
        }
      }
    }
  }
  cout << max(dp[(1 << n) - 1][0], dp[(1 << n) - 1][1]) << '\n';
  return 0;
}
