#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
template <class T>
bool uin(T& a, const T& b) {
  return a >= b ? a = b, 1 : 0;
}
template <class T>
bool uax(T& a, const T& b) {
  return a <= b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
constexpr int inf = 1e9;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> a(n);
  vector<int> sum(n), low(n);
  for (auto i = (0); i < (n); ++i) {
    string s;
    cin >> s;
    a[i].assign(int((s).size()) + 1, 0);
    for (auto j = (0); j < (int((s).size())); ++j) {
      sum[i] += s[j] == '(' ? +1 : -1;
      if (uax(low[i], -sum[i])) ++a[i][low[i]];
    }
  }
  vector<int> dp(1 << n, -inf);
  dp[0] = 0;
  int ans = 0;
  for (auto S = (0); S < (1 << n); ++S) {
    uax(ans, dp[S]);
    int count = 0;
    for (auto i = (0); i < (n); ++i)
      if (S >> i & 1) count += sum[i];
    for (auto i = (0); i < (n); ++i) {
      if (S >> i & 1) continue;
      int del = 0 <= count && count < int((a[i]).size()) ? a[i][count] : 0;
      uax(count < low[i] ? ans : dp[S | 1 << i], dp[S] + del);
    }
  }
  cout << ans << '\n';
}

### Prompt

In Cpp, your task is to solve 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;
using ll = long long;
using ld = long double;
template <class T>
bool uin(T& a, const T& b) {
  return a >= b ? a = b, 1 : 0;
}
template <class T>
bool uax(T& a, const T& b) {
  return a <= b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
constexpr int inf = 1e9;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> a(n);
  vector<int> sum(n), low(n);
  for (auto i = (0); i < (n); ++i) {
    string s;
    cin >> s;
    a[i].assign(int((s).size()) + 1, 0);
    for (auto j = (0); j < (int((s).size())); ++j) {
      sum[i] += s[j] == '(' ? +1 : -1;
      if (uax(low[i], -sum[i])) ++a[i][low[i]];
    }
  }
  vector<int> dp(1 << n, -inf);
  dp[0] = 0;
  int ans = 0;
  for (auto S = (0); S < (1 << n); ++S) {
    uax(ans, dp[S]);
    int count = 0;
    for (auto i = (0); i < (n); ++i)
      if (S >> i & 1) count += sum[i];
    for (auto i = (0); i < (n); ++i) {
      if (S >> i & 1) continue;
      int del = 0 <= count && count < int((a[i]).size()) ? a[i][count] : 0;
      uax(count < low[i] ? ans : dp[S | 1 << i], dp[S] + del);
    }
  }
  cout << ans << '\n';
}

```