#include <bits/stdc++.h>
using namespace std;
int n;
int NN = 20;
string s[20];
int cnt[20][(int)1e6];
vector<int> tot(20, 0);
vector<int> ts((1 << NN), 0);
vector<int> vis((1 << NN), 0);
vector<int> dp((1 << NN), 0);
vector<int> lst((1 << NN), 0);
int ans = 0;
int fun(int x) {
  if (vis[x]) return dp[x];
  vis[x]++;
  if (ts[x] < 0) return 0;
  int cmax = 0;
  for (int i = 0; i < n; i++) {
    if (((1 << i) & x) == 0) {
      int y = cnt[i][(int)5e5 - ts[x]];
      if (lst[(1 << i)] + ts[x] >= 0) y += fun(x | (1 << i));
      cmax = max(cmax, y);
    }
  }
  dp[x] = cmax;
  return cmax;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    int curr = 0;
    bool less = false;
    int least = 0;
    for (auto j : s[i]) {
      if (j == '(')
        curr++;
      else
        curr--;
      if (curr <= 0 && least >= curr) cnt[i][(int)5e5 + curr]++;
      if (curr < 0) {
        least = min(least, curr);
      }
    }
    lst[(1 << i)] = least;
    tot[i] = curr;
  }
  int N = (1 << n);
  for (int i = 1; i < N; i++) {
    int a = 0;
    for (int j = 0; j < 20; j++) {
      if (((1 << j) & i) != 0) a += tot[j];
    }
    ts[i] = a;
  }
  fun(0);
  cout << dp[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>
using namespace std;
int n;
int NN = 20;
string s[20];
int cnt[20][(int)1e6];
vector<int> tot(20, 0);
vector<int> ts((1 << NN), 0);
vector<int> vis((1 << NN), 0);
vector<int> dp((1 << NN), 0);
vector<int> lst((1 << NN), 0);
int ans = 0;
int fun(int x) {
  if (vis[x]) return dp[x];
  vis[x]++;
  if (ts[x] < 0) return 0;
  int cmax = 0;
  for (int i = 0; i < n; i++) {
    if (((1 << i) & x) == 0) {
      int y = cnt[i][(int)5e5 - ts[x]];
      if (lst[(1 << i)] + ts[x] >= 0) y += fun(x | (1 << i));
      cmax = max(cmax, y);
    }
  }
  dp[x] = cmax;
  return cmax;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    int curr = 0;
    bool less = false;
    int least = 0;
    for (auto j : s[i]) {
      if (j == '(')
        curr++;
      else
        curr--;
      if (curr <= 0 && least >= curr) cnt[i][(int)5e5 + curr]++;
      if (curr < 0) {
        least = min(least, curr);
      }
    }
    lst[(1 << i)] = least;
    tot[i] = curr;
  }
  int N = (1 << n);
  for (int i = 1; i < N; i++) {
    int a = 0;
    for (int j = 0; j < 20; j++) {
      if (((1 << j) & i) != 0) a += tot[j];
    }
    ts[i] = a;
  }
  fun(0);
  cout << dp[0];
}

```