#include <bits/stdc++.h>
using namespace std;
const int inf = 1e6;
const int oo = 1e9;
const int smax = 4e5;
int n, m, num, d[21], c[21], pow2[21], f[1048580][2], sum[1048580];
bool kt;
string s;
pair<int, int> goes[21][400001], temp, tam;
vector<int> v[smax];
pair<int, int> go(int i, int flag, int vaule) {
  if (flag) return {0, 1};
  return goes[i][vaule];
}
void maximize(int *x, int y) {
  if (*x < y) *x = y;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(f, -1, sizeof(f));
  pow2[0] = 1;
  for (int i = 1; i <= 20; i++) pow2[i] = pow2[i - 1] * 2;
  cin >> n;
  for (int j = 1; j < pow2[n]; j++) f[j][0] = f[j][1] = -oo;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= smax; j++) v[j].clear();
    cin >> s;
    m = s.size();
    c[i] = d[i] = 0;
    kt = true;
    for (int j = 0; j < m; j++) {
      if (s[j] == '(')
        d[i]++;
      else
        d[i]--;
      if (d[i] < c[i]) c[i] = d[i];
      if (d[i] <= 0) v[-d[i]].push_back(j);
    }
    for (int j = 0; j <= smax; j++) {
      if (v[j].empty())
        num = 0;
      else {
        if (!v[j + 1].empty())
          num =
              upper_bound(v[j].begin(), v[j].end(), v[j + 1][0]) - v[j].begin();
        else
          num = v[j].size();
      }
      if (c[i] + j < 0)
        goes[i][j] = {num, 1};
      else
        goes[i][j] = {num, 0};
    }
    for (int j = 1; j < pow2[n]; j++)
      if ((j >> (i - 1)) & 1) sum[j] += d[i];
  }
  f[0][0] = sum[0] = 0;
  f[0][1] = -oo;
  for (int i = 0; i < pow2[n]; i++) {
    for (int j = 1; j <= n; j++)
      if (!((i >> (j - 1)) & 1)) {
        if (f[i][1] != -oo) maximize(&f[i ^ pow2[j - 1]][1], f[i][1]);
        if (f[i][0] == -oo) continue;
        temp = go(j, 0, sum[i]);
        maximize(&f[i ^ pow2[j - 1]][temp.second], f[i][0] + temp.first);
      }
  }
  cout << max(f[pow2[n] - 1][0], f[pow2[n] - 1][1]);
  return 0;
}

### 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;
const int inf = 1e6;
const int oo = 1e9;
const int smax = 4e5;
int n, m, num, d[21], c[21], pow2[21], f[1048580][2], sum[1048580];
bool kt;
string s;
pair<int, int> goes[21][400001], temp, tam;
vector<int> v[smax];
pair<int, int> go(int i, int flag, int vaule) {
  if (flag) return {0, 1};
  return goes[i][vaule];
}
void maximize(int *x, int y) {
  if (*x < y) *x = y;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(f, -1, sizeof(f));
  pow2[0] = 1;
  for (int i = 1; i <= 20; i++) pow2[i] = pow2[i - 1] * 2;
  cin >> n;
  for (int j = 1; j < pow2[n]; j++) f[j][0] = f[j][1] = -oo;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= smax; j++) v[j].clear();
    cin >> s;
    m = s.size();
    c[i] = d[i] = 0;
    kt = true;
    for (int j = 0; j < m; j++) {
      if (s[j] == '(')
        d[i]++;
      else
        d[i]--;
      if (d[i] < c[i]) c[i] = d[i];
      if (d[i] <= 0) v[-d[i]].push_back(j);
    }
    for (int j = 0; j <= smax; j++) {
      if (v[j].empty())
        num = 0;
      else {
        if (!v[j + 1].empty())
          num =
              upper_bound(v[j].begin(), v[j].end(), v[j + 1][0]) - v[j].begin();
        else
          num = v[j].size();
      }
      if (c[i] + j < 0)
        goes[i][j] = {num, 1};
      else
        goes[i][j] = {num, 0};
    }
    for (int j = 1; j < pow2[n]; j++)
      if ((j >> (i - 1)) & 1) sum[j] += d[i];
  }
  f[0][0] = sum[0] = 0;
  f[0][1] = -oo;
  for (int i = 0; i < pow2[n]; i++) {
    for (int j = 1; j <= n; j++)
      if (!((i >> (j - 1)) & 1)) {
        if (f[i][1] != -oo) maximize(&f[i ^ pow2[j - 1]][1], f[i][1]);
        if (f[i][0] == -oo) continue;
        temp = go(j, 0, sum[i]);
        maximize(&f[i ^ pow2[j - 1]][temp.second], f[i][0] + temp.first);
      }
  }
  cout << max(f[pow2[n] - 1][0], f[pow2[n] - 1][1]);
  return 0;
}

```