#include <bits/stdc++.h>
using namespace std;
int n, dp[2][1048576], cnt[1048576], maxn[23];
string s[23];
int mp[23][800003];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    int qaq = 0;
    for (int j = 0; j < s[i].size(); j++)
      if (s[i][j] == '(')
        qaq++;
      else
        qaq--, maxn[i] = min(maxn[i], qaq);
    cnt[1 << i] = qaq;
  }
  for (int i = 0; i < (1 << n); i++)
    if (__builtin_popcount(i) > 1)
      for (int j = 0; j < n; j++)
        if (i & (1 << j)) cnt[i] += cnt[1 << j];
  for (int i = 0; i < n; i++) {
    int minn = 0, qaq = 0, CNT = 0;
    for (int j = 0; j < s[i].size(); j++) {
      if (s[i][j] == '(')
        qaq++;
      else
        qaq--;
      if (qaq < minn) CNT = 0;
      minn = min(minn, qaq);
      if (qaq == minn) CNT++;
      mp[i][-minn] = CNT;
    }
  }
  for (int i = 0; i < (1 << n); i++) dp[0][i] = -19260817, dp[1][i] = -19260817;
  dp[1][0] = 0;
  for (int i = 1; i < (1 << n); i++)
    for (int j = 0; j < n; j++)
      if (i & (1 << j)) {
        dp[0][i] = max(dp[0][i], dp[0][i ^ (1 << j)]);
        if (dp[1][i ^ (1 << j)] >= 0)
          dp[(maxn[j] + cnt[i ^ (1 << j)]) >= 0][i] =
              max(dp[(maxn[j] + cnt[i ^ (1 << j)]) >= 0][i],
                  dp[1][i ^ (1 << j)] + mp[j][cnt[i ^ (1 << j)]]);
      }
  cout << max(dp[0][(1 << n) - 1], dp[1][(1 << n) - 1]);
}

### Prompt

Your task is to create 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, dp[2][1048576], cnt[1048576], maxn[23];
string s[23];
int mp[23][800003];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    int qaq = 0;
    for (int j = 0; j < s[i].size(); j++)
      if (s[i][j] == '(')
        qaq++;
      else
        qaq--, maxn[i] = min(maxn[i], qaq);
    cnt[1 << i] = qaq;
  }
  for (int i = 0; i < (1 << n); i++)
    if (__builtin_popcount(i) > 1)
      for (int j = 0; j < n; j++)
        if (i & (1 << j)) cnt[i] += cnt[1 << j];
  for (int i = 0; i < n; i++) {
    int minn = 0, qaq = 0, CNT = 0;
    for (int j = 0; j < s[i].size(); j++) {
      if (s[i][j] == '(')
        qaq++;
      else
        qaq--;
      if (qaq < minn) CNT = 0;
      minn = min(minn, qaq);
      if (qaq == minn) CNT++;
      mp[i][-minn] = CNT;
    }
  }
  for (int i = 0; i < (1 << n); i++) dp[0][i] = -19260817, dp[1][i] = -19260817;
  dp[1][0] = 0;
  for (int i = 1; i < (1 << n); i++)
    for (int j = 0; j < n; j++)
      if (i & (1 << j)) {
        dp[0][i] = max(dp[0][i], dp[0][i ^ (1 << j)]);
        if (dp[1][i ^ (1 << j)] >= 0)
          dp[(maxn[j] + cnt[i ^ (1 << j)]) >= 0][i] =
              max(dp[(maxn[j] + cnt[i ^ (1 << j)]) >= 0][i],
                  dp[1][i ^ (1 << j)] + mp[j][cnt[i ^ (1 << j)]]);
      }
  cout << max(dp[0][(1 << n) - 1], dp[1][(1 << n) - 1]);
}

```