#include <bits/stdc++.h>
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  std::vector<std::string> s(n);
  for (int i = 0; i < n; ++i) std::cin >> s[i];
  std::vector<int> len(n), low(n), h(n);
  std::vector<std::vector<int>> tax(n);
  for (int i = 0; i < n; ++i) {
    len[i] = s[i].length();
    h[i] = 0;
    low[i] = 0;
    tax[i].resize(len[i] + 1);
    for (char c : s[i]) {
      h[i] += c == '(' ? 1 : -1;
      if (h[i] <= low[i]) {
        low[i] = std::min(low[i], h[i]);
        ++tax[i][h[i] + len[i]];
      }
    }
  }
  std::vector<int> f(1 << n), pre(1 << n);
  int ans = 0;
  for (int s = 1; s < (1 << n); ++s) {
    f[s] = -1;
    for (int i = 0; i < n; ++i) {
      if ((s >> i) & 1) {
        pre[s] += h[i];
        int t = s ^ (1 << i);
        if (f[t] == -1) continue;
        int val = f[t];
        if (pre[t] >= 0 && pre[t] <= len[i]) val += tax[i][len[i] - pre[t]];
        ans = std::max(ans, val);
        if (pre[t] + low[i] >= 0) f[s] = std::max(f[s], val);
      }
    }
  }
  std::cout << ans << '\n';
  return 0;
}

### Prompt

Generate 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>
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  std::vector<std::string> s(n);
  for (int i = 0; i < n; ++i) std::cin >> s[i];
  std::vector<int> len(n), low(n), h(n);
  std::vector<std::vector<int>> tax(n);
  for (int i = 0; i < n; ++i) {
    len[i] = s[i].length();
    h[i] = 0;
    low[i] = 0;
    tax[i].resize(len[i] + 1);
    for (char c : s[i]) {
      h[i] += c == '(' ? 1 : -1;
      if (h[i] <= low[i]) {
        low[i] = std::min(low[i], h[i]);
        ++tax[i][h[i] + len[i]];
      }
    }
  }
  std::vector<int> f(1 << n), pre(1 << n);
  int ans = 0;
  for (int s = 1; s < (1 << n); ++s) {
    f[s] = -1;
    for (int i = 0; i < n; ++i) {
      if ((s >> i) & 1) {
        pre[s] += h[i];
        int t = s ^ (1 << i);
        if (f[t] == -1) continue;
        int val = f[t];
        if (pre[t] >= 0 && pre[t] <= len[i]) val += tax[i][len[i] - pre[t]];
        ans = std::max(ans, val);
        if (pre[t] + low[i] >= 0) f[s] = std::max(f[s], val);
      }
    }
  }
  std::cout << ans << '\n';
  return 0;
}

```