#include <bits/stdc++.h>
using namespace std;
auto fio = (ios::sync_with_stdio(0), cin.tie(0), cout.tie(0));
int32_t main() {
  int n;
  cin >> n;
  vector<string> s(n);
  for (int i = 0; i < n; i++) cin >> s[i];
  vector<vector<int>> pfx(n), cnt(n);
  vector<int> all(n);
  for (int i = 0; i < n; i++) {
    int sz = s[i].size();
    int SH = sz + 2;
    cnt[i].resize(2 * sz + 4);
    pfx[i].resize(2 * sz + 4, -1);
    int bal = 0;
    for (int j = 0; j < sz; j++) {
      bal += s[i][j] == '(' ? +1 : -1;
      if (pfx[i][bal + SH] == -1) {
        pfx[i][bal + SH] = cnt[i][bal + 1 + SH];
      }
      cnt[i][bal + SH]++;
    }
    all[i] = bal;
  }
  auto smax = [](int& a, int b) { return a = max(a, b); };
  int ans = 0;
  vector<int> dp(1 << n, -1);
  dp[0] = 0;
  for (int msk = 0; msk < 1 << n; msk++) {
    if (dp[msk] == -1) continue;
    int bal = 0;
    for (int i = 0; i < n; i++)
      if (msk >> i & 1) bal += all[i];
    if (bal < 0) continue;
    for (int i = 0; i < n; i++) {
      int sz = s[i].size();
      int SH = sz + 2;
      if (~msk >> i & 1) {
        if (bal > sz)
          smax(dp[msk | 1 << i], dp[msk]);
        else {
          if (pfx[i][-bal - 1 + SH] != -1) {
            smax(ans, dp[msk] + pfx[i][-bal - 1 + SH]);
          } else {
            smax(dp[msk | 1 << i], dp[msk] + cnt[i][-bal + SH]);
          }
        }
      }
    }
    smax(ans, dp[msk]);
  }
  cout << ans << endl;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
auto fio = (ios::sync_with_stdio(0), cin.tie(0), cout.tie(0));
int32_t main() {
  int n;
  cin >> n;
  vector<string> s(n);
  for (int i = 0; i < n; i++) cin >> s[i];
  vector<vector<int>> pfx(n), cnt(n);
  vector<int> all(n);
  for (int i = 0; i < n; i++) {
    int sz = s[i].size();
    int SH = sz + 2;
    cnt[i].resize(2 * sz + 4);
    pfx[i].resize(2 * sz + 4, -1);
    int bal = 0;
    for (int j = 0; j < sz; j++) {
      bal += s[i][j] == '(' ? +1 : -1;
      if (pfx[i][bal + SH] == -1) {
        pfx[i][bal + SH] = cnt[i][bal + 1 + SH];
      }
      cnt[i][bal + SH]++;
    }
    all[i] = bal;
  }
  auto smax = [](int& a, int b) { return a = max(a, b); };
  int ans = 0;
  vector<int> dp(1 << n, -1);
  dp[0] = 0;
  for (int msk = 0; msk < 1 << n; msk++) {
    if (dp[msk] == -1) continue;
    int bal = 0;
    for (int i = 0; i < n; i++)
      if (msk >> i & 1) bal += all[i];
    if (bal < 0) continue;
    for (int i = 0; i < n; i++) {
      int sz = s[i].size();
      int SH = sz + 2;
      if (~msk >> i & 1) {
        if (bal > sz)
          smax(dp[msk | 1 << i], dp[msk]);
        else {
          if (pfx[i][-bal - 1 + SH] != -1) {
            smax(ans, dp[msk] + pfx[i][-bal - 1 + SH]);
          } else {
            smax(dp[msk | 1 << i], dp[msk] + cnt[i][-bal + SH]);
          }
        }
      }
    }
    smax(ans, dp[msk]);
  }
  cout << ans << endl;
}

```