#include <bits/stdc++.h>
using namespace std;
int n, m, len[21], ans, f[21][800011], w[21], ss[1100001], val[21];
char ch[21][400011];
bool fl[21][800011];
int main() {
  val[0] = 1;
  for (int i = 1; i <= 20; i++) val[i] = val[i - 1] + val[i - 1];
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", ch[i] + 1);
    len[i] = strlen(ch[i] + 1);
    for (int j = 1; j <= len[i]; j++) {
      if (ch[i][j] == '(')
        w[i]++;
      else
        w[i]--;
      int s = 400000 - w[i] - 1;
      while (s >= 0) {
        if (fl[i][s]) break;
        fl[i][s] = 1;
        s--;
      }
      if (!fl[i][400000 - w[i]]) f[i][400000 - w[i]]++;
    }
  }
  for (int i = 1; i < val[n]; i++) ss[i] = -1000000000;
  for (int i = 0; i < val[n]; i++) {
    ans = max(ans, ss[i]);
    int s = 400000;
    for (int j = 1; j <= n; j++)
      if (val[j - 1] & i) s = s + w[j];
    for (int j = 1; j <= n; j++) {
      if (val[j - 1] & i) continue;
      if (!fl[j][s]) {
        ss[i | val[j - 1]] = max(ss[i | val[j - 1]], ss[i] + f[j][s]);
      } else {
        ans = max(ans, ss[i] + f[j][s]);
      }
    }
  }
  printf("%d\n", ans);
}

### Prompt

Create a solution in CPP for 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, m, len[21], ans, f[21][800011], w[21], ss[1100001], val[21];
char ch[21][400011];
bool fl[21][800011];
int main() {
  val[0] = 1;
  for (int i = 1; i <= 20; i++) val[i] = val[i - 1] + val[i - 1];
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", ch[i] + 1);
    len[i] = strlen(ch[i] + 1);
    for (int j = 1; j <= len[i]; j++) {
      if (ch[i][j] == '(')
        w[i]++;
      else
        w[i]--;
      int s = 400000 - w[i] - 1;
      while (s >= 0) {
        if (fl[i][s]) break;
        fl[i][s] = 1;
        s--;
      }
      if (!fl[i][400000 - w[i]]) f[i][400000 - w[i]]++;
    }
  }
  for (int i = 1; i < val[n]; i++) ss[i] = -1000000000;
  for (int i = 0; i < val[n]; i++) {
    ans = max(ans, ss[i]);
    int s = 400000;
    for (int j = 1; j <= n; j++)
      if (val[j - 1] & i) s = s + w[j];
    for (int j = 1; j <= n; j++) {
      if (val[j - 1] & i) continue;
      if (!fl[j][s]) {
        ss[i | val[j - 1]] = max(ss[i | val[j - 1]], ss[i] + f[j][s]);
      } else {
        ans = max(ans, ss[i] + f[j][s]);
      }
    }
  }
  printf("%d\n", ans);
}

```