#include <bits/stdc++.h>
using namespace std;
const int N = 20;
const int M = 400005;
string s[N];
int n, minH[N], sum[N], f[1 << N], freq[N][M];
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    int cur = 0;
    vector<pair<int, int>> vec;
    for (int j = 0; j < s[i].size(); ++j) {
      char c = s[i][j];
      if (c == '(')
        ++cur;
      else
        --cur;
      minH[i] = min(minH[i], cur);
      vec.emplace_back(cur, -j);
    }
    sum[i] = cur;
    sort(vec.begin(), vec.end());
    int bound = s[i].size();
    for (auto [val, pos] : vec) {
      pos = -pos, val = -val;
      if (pos < bound and val >= 0) ++freq[i][val];
      bound = min(bound, pos);
    }
  }
  for (int mask = (1 << n) - 1; mask >= 0; --mask) {
    int tot = 0;
    for (int i = 0; i < n; ++i)
      if (mask & 1 << i) tot += sum[i];
    if (tot < 0) continue;
    for (int i = 0; i < n; ++i)
      if (~mask & 1 << i) {
        int h = minH[i], cur = sum[i], here = 0;
        if (tot + h < 0) {
          f[mask] = max(f[mask], freq[i][tot]);
        } else {
          f[mask] = max(f[mask], freq[i][tot] + f[mask | 1 << i]);
        }
      }
  }
  cout << f[0] << '\n';
  return 0;
}

### 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;
const int N = 20;
const int M = 400005;
string s[N];
int n, minH[N], sum[N], f[1 << N], freq[N][M];
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    int cur = 0;
    vector<pair<int, int>> vec;
    for (int j = 0; j < s[i].size(); ++j) {
      char c = s[i][j];
      if (c == '(')
        ++cur;
      else
        --cur;
      minH[i] = min(minH[i], cur);
      vec.emplace_back(cur, -j);
    }
    sum[i] = cur;
    sort(vec.begin(), vec.end());
    int bound = s[i].size();
    for (auto [val, pos] : vec) {
      pos = -pos, val = -val;
      if (pos < bound and val >= 0) ++freq[i][val];
      bound = min(bound, pos);
    }
  }
  for (int mask = (1 << n) - 1; mask >= 0; --mask) {
    int tot = 0;
    for (int i = 0; i < n; ++i)
      if (mask & 1 << i) tot += sum[i];
    if (tot < 0) continue;
    for (int i = 0; i < n; ++i)
      if (~mask & 1 << i) {
        int h = minH[i], cur = sum[i], here = 0;
        if (tot + h < 0) {
          f[mask] = max(f[mask], freq[i][tot]);
        } else {
          f[mask] = max(f[mask], freq[i][tot] + f[mask | 1 << i]);
        }
      }
  }
  cout << f[0] << '\n';
  return 0;
}

```