#include <bits/stdc++.h>
using namespace std;
const int P = 1 << 20;
int bad[21];
int dp[P][2], le[P];
int main() {
  int n;
  cin >> n;
  map<int, int> f[20];
  memset(dp, -0x3f, sizeof(dp));
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    int sum = 0, mn = 0, cnt = 0;
    for (int j = 0; j < (int)s.size(); ++j) {
      if (s[j] == '(')
        ++sum;
      else
        --sum;
      mn = min(mn, sum);
      if (sum == mn) ++f[i][-sum];
      if (mn == 0 && sum == 0) ++cnt;
    }
    bad[i] = mn;
    le[1 << i] = sum;
    if (mn < 0)
      dp[1 << i][0] = cnt;
    else
      dp[1 << i][1] = cnt;
  }
  int p = (1 << n);
  for (int mask = 1; mask < p; ++mask) {
    for (int bit = 0; bit < n; ++bit) {
      if (mask & (1 << bit)) continue;
      dp[mask | (1 << bit)][0] = max(dp[mask | (1 << bit)][0], dp[mask][0]);
      int aux = 0;
      if (f[bit].count(le[mask])) aux = f[bit][le[mask]];
      if (le[mask] + bad[bit] >= 0)
        dp[mask | (1 << bit)][1] =
            max(dp[mask | (1 << bit)][1], dp[mask][1] + aux);
      else
        dp[mask | (1 << bit)][0] =
            max(dp[mask | (1 << bit)][0], dp[mask][1] + aux);
      le[mask | (1 << bit)] = le[mask] + le[1 << bit];
    }
  }
  printf("%d", max(dp[p - 1][0], dp[p - 1][1]));
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 P = 1 << 20;
int bad[21];
int dp[P][2], le[P];
int main() {
  int n;
  cin >> n;
  map<int, int> f[20];
  memset(dp, -0x3f, sizeof(dp));
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    int sum = 0, mn = 0, cnt = 0;
    for (int j = 0; j < (int)s.size(); ++j) {
      if (s[j] == '(')
        ++sum;
      else
        --sum;
      mn = min(mn, sum);
      if (sum == mn) ++f[i][-sum];
      if (mn == 0 && sum == 0) ++cnt;
    }
    bad[i] = mn;
    le[1 << i] = sum;
    if (mn < 0)
      dp[1 << i][0] = cnt;
    else
      dp[1 << i][1] = cnt;
  }
  int p = (1 << n);
  for (int mask = 1; mask < p; ++mask) {
    for (int bit = 0; bit < n; ++bit) {
      if (mask & (1 << bit)) continue;
      dp[mask | (1 << bit)][0] = max(dp[mask | (1 << bit)][0], dp[mask][0]);
      int aux = 0;
      if (f[bit].count(le[mask])) aux = f[bit][le[mask]];
      if (le[mask] + bad[bit] >= 0)
        dp[mask | (1 << bit)][1] =
            max(dp[mask | (1 << bit)][1], dp[mask][1] + aux);
      else
        dp[mask | (1 << bit)][0] =
            max(dp[mask | (1 << bit)][0], dp[mask][1] + aux);
      le[mask | (1 << bit)] = le[mask] + le[1 << bit];
    }
  }
  printf("%d", max(dp[p - 1][0], dp[p - 1][1]));
  return 0;
}

```