#include <bits/stdc++.h>
using namespace std;
template <class _F, class _S>
ostream &operator<<(ostream &out, const pair<_F, _S> &_pair) {
  return out << "(" << _pair.first << "," << _pair.second << ")";
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &_vec) {
  out << "[";
  for (int _vecp = 0; _vecp < _vec.size(); ++_vecp)
    out << _vec[_vecp] << ",]"[_vecp + 1 == _vec.size()];
  return out;
}
void debug() {}
template <class T, class... U>
void debug(const T &head, const U &...tail) {
  cerr << head << (sizeof...(tail) ? " " : "\n"), debug(tail...);
}
template <class... T>
void D(const T &...parm) {
  cerr << ">>> ", debug(parm...);
}
const int mx = 4e5 + 10;
const long long mod = 998244353;
mt19937_64 mtrand(time(0));
int n, m, T;
string s[20];
int b[20];
int cnt[1 << 20];
vector<int> vec[20];
unordered_map<int, int> mp[20], mp2[20];
int dp[1 << 20][2];
int sk[mx], tot;
void solve(int id) {
  for (int i = 0; i < s[id].size(); ++i) {
    if (s[id][i] == '(')
      ++b[id];
    else
      --b[id];
    vec[id].push_back(b[id]);
    ++mp2[id][b[id]];
    if (!mp[id].count(b[id])) mp[id][b[id]] = mp2[id][b[id] + 1];
  }
}
void run_case() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    solve(i);
  }
  int up = 1 << n;
  for (int i = 0; i < up; ++i) {
    for (int j = 0; j < n; ++j) {
      if ((i >> j) & 1) cnt[i] += b[j];
    }
    dp[i][0] = dp[i][1] = -0x3f3f3f3f;
  }
  dp[0][0] = 0;
  for (int i = 1; i < up; ++i) {
    for (int j = 0; j < n; ++j) {
      if ((i >> j) & 1) {
        int st = i & (up - (1 << j) - 1);
        if (mp[j].count(-cnt[st] - 1)) {
          dp[i][1] = max(dp[i][1], dp[st][0] + mp[j][-cnt[st] - 1]);
        } else {
          dp[i][0] = max(dp[i][0], dp[st][0] + mp2[j][-cnt[st]]);
        }
        dp[i][1] = max(dp[i][1], dp[st][1]);
      }
    }
  }
  long long ans = max(dp[up - 1][0], dp[up - 1][1]);
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  T = 1;
  for (int i = 1; i <= T; ++i) run_case();
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
template <class _F, class _S>
ostream &operator<<(ostream &out, const pair<_F, _S> &_pair) {
  return out << "(" << _pair.first << "," << _pair.second << ")";
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &_vec) {
  out << "[";
  for (int _vecp = 0; _vecp < _vec.size(); ++_vecp)
    out << _vec[_vecp] << ",]"[_vecp + 1 == _vec.size()];
  return out;
}
void debug() {}
template <class T, class... U>
void debug(const T &head, const U &...tail) {
  cerr << head << (sizeof...(tail) ? " " : "\n"), debug(tail...);
}
template <class... T>
void D(const T &...parm) {
  cerr << ">>> ", debug(parm...);
}
const int mx = 4e5 + 10;
const long long mod = 998244353;
mt19937_64 mtrand(time(0));
int n, m, T;
string s[20];
int b[20];
int cnt[1 << 20];
vector<int> vec[20];
unordered_map<int, int> mp[20], mp2[20];
int dp[1 << 20][2];
int sk[mx], tot;
void solve(int id) {
  for (int i = 0; i < s[id].size(); ++i) {
    if (s[id][i] == '(')
      ++b[id];
    else
      --b[id];
    vec[id].push_back(b[id]);
    ++mp2[id][b[id]];
    if (!mp[id].count(b[id])) mp[id][b[id]] = mp2[id][b[id] + 1];
  }
}
void run_case() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    solve(i);
  }
  int up = 1 << n;
  for (int i = 0; i < up; ++i) {
    for (int j = 0; j < n; ++j) {
      if ((i >> j) & 1) cnt[i] += b[j];
    }
    dp[i][0] = dp[i][1] = -0x3f3f3f3f;
  }
  dp[0][0] = 0;
  for (int i = 1; i < up; ++i) {
    for (int j = 0; j < n; ++j) {
      if ((i >> j) & 1) {
        int st = i & (up - (1 << j) - 1);
        if (mp[j].count(-cnt[st] - 1)) {
          dp[i][1] = max(dp[i][1], dp[st][0] + mp[j][-cnt[st] - 1]);
        } else {
          dp[i][0] = max(dp[i][0], dp[st][0] + mp2[j][-cnt[st]]);
        }
        dp[i][1] = max(dp[i][1], dp[st][1]);
      }
    }
  }
  long long ans = max(dp[up - 1][0], dp[up - 1][1]);
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  T = 1;
  for (int i = 1; i <= T; ++i) run_case();
}

```