#include <bits/stdc++.h>
using namespace std;
const int MAX = 4e5 + 5;
int summ[20];
int mmin[20];
int maskSum[1 << 20];
int cnt[MAX][20];
int dp[1 << 20];
int sol = 0;
string arr[20];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < (1 << n); i++) dp[i] = -1;
  for (int i = 0; i < n; i++) {
    string s = arr[i];
    for (int j = 0; j < s.size(); j++) {
      if (s[j] == '(')
        summ[i]++;
      else
        summ[i]--;
      mmin[i] = min(mmin[i], summ[i]);
      if (summ[i] == mmin[i]) cnt[-mmin[i]][i]++;
    }
  }
  dp[0] = 0;
  for (int mask = 0; mask < (1 << n); mask++) {
    if (dp[mask] < 0) continue;
    for (int j = 0; j < n; j++) {
      if (mask & (1 << j)) continue;
      int mask2 = mask | (1 << j);
      int val = dp[mask] + cnt[maskSum[mask]][j];
      sol = max(sol, val);
      maskSum[mask2] = maskSum[mask] + summ[j];
      if (maskSum[mask] + mmin[j] >= 0) {
        dp[mask2] = max(dp[mask2], val);
      }
    }
  }
  cout << sol << "\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>
using namespace std;
const int MAX = 4e5 + 5;
int summ[20];
int mmin[20];
int maskSum[1 << 20];
int cnt[MAX][20];
int dp[1 << 20];
int sol = 0;
string arr[20];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < (1 << n); i++) dp[i] = -1;
  for (int i = 0; i < n; i++) {
    string s = arr[i];
    for (int j = 0; j < s.size(); j++) {
      if (s[j] == '(')
        summ[i]++;
      else
        summ[i]--;
      mmin[i] = min(mmin[i], summ[i]);
      if (summ[i] == mmin[i]) cnt[-mmin[i]][i]++;
    }
  }
  dp[0] = 0;
  for (int mask = 0; mask < (1 << n); mask++) {
    if (dp[mask] < 0) continue;
    for (int j = 0; j < n; j++) {
      if (mask & (1 << j)) continue;
      int mask2 = mask | (1 << j);
      int val = dp[mask] + cnt[maskSum[mask]][j];
      sol = max(sol, val);
      maskSum[mask2] = maskSum[mask] + summ[j];
      if (maskSum[mask] + mmin[j] >= 0) {
        dp[mask2] = max(dp[mask2], val);
      }
    }
  }
  cout << sol << "\n";
  return 0;
}

```