#include <bits/stdc++.h>
using namespace std;
int N;
string S[20];
int cnt[20], mc[20], ap[20];
vector<pair<int, int> > vis[20];
int sum[1 << 20];
int dp[1 << 20];
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> S[i];
    mc[i] = 0;
    for (char c : S[i]) {
      cnt[i] += c == '(' ? 1 : -1;
      if (mc[i] > cnt[i]) {
        vis[i].push_back(make_pair(mc[i], ap[i]));
        ap[i] = 0;
        mc[i] = cnt[i];
      }
      if (mc[i] == cnt[i]) ap[i]++;
    }
    vis[i].push_back(make_pair(mc[i], ap[i]));
    sort(vis[i].begin(), vis[i].end());
  }
  for (int i = 0; i < 1 << N; i++) dp[i] = -1;
  dp[0] = 0;
  int ans = 0;
  for (int i = 0; i < 1 << N; i++)
    if (dp[i] >= 0) {
      if (ans < dp[i]) ans = dp[i];
      int now = sum[i];
      for (int j = 0; j < N; j++)
        if (!(i >> j & 1)) {
          auto it =
              lower_bound(vis[j].begin(), vis[j].end(), make_pair(-now, 0));
          if (it != vis[j].end() && it->first == -now) {
            ans = max(ans, dp[i] + it->second);
          }
          if (now + mc[j] < 0) continue;
          int nxt = dp[i];
          if (now + mc[j] == 0) nxt += ap[j];
          int ni = i | 1 << j;
          sum[ni] = now + cnt[j];
          if (dp[ni] < nxt) dp[ni] = nxt;
        }
    }
  cout << ans << endl;
}

### Prompt

Please formulate a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int N;
string S[20];
int cnt[20], mc[20], ap[20];
vector<pair<int, int> > vis[20];
int sum[1 << 20];
int dp[1 << 20];
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> S[i];
    mc[i] = 0;
    for (char c : S[i]) {
      cnt[i] += c == '(' ? 1 : -1;
      if (mc[i] > cnt[i]) {
        vis[i].push_back(make_pair(mc[i], ap[i]));
        ap[i] = 0;
        mc[i] = cnt[i];
      }
      if (mc[i] == cnt[i]) ap[i]++;
    }
    vis[i].push_back(make_pair(mc[i], ap[i]));
    sort(vis[i].begin(), vis[i].end());
  }
  for (int i = 0; i < 1 << N; i++) dp[i] = -1;
  dp[0] = 0;
  int ans = 0;
  for (int i = 0; i < 1 << N; i++)
    if (dp[i] >= 0) {
      if (ans < dp[i]) ans = dp[i];
      int now = sum[i];
      for (int j = 0; j < N; j++)
        if (!(i >> j & 1)) {
          auto it =
              lower_bound(vis[j].begin(), vis[j].end(), make_pair(-now, 0));
          if (it != vis[j].end() && it->first == -now) {
            ans = max(ans, dp[i] + it->second);
          }
          if (now + mc[j] < 0) continue;
          int nxt = dp[i];
          if (now + mc[j] == 0) nxt += ap[j];
          int ni = i | 1 << j;
          sum[ni] = now + cnt[j];
          if (dp[ni] < nxt) dp[ni] = nxt;
        }
    }
  cout << ans << endl;
}

```