#include <bits/stdc++.h>
using namespace std;
int n;
const int MAXN = 4e5 + 7;
char s[MAXN];
int g[20][MAXN], w[20], mn[20], f[1 << 20], W[1 << 20], ans;
int main() {
  memset(f, -0x3f, sizeof(f));
  f[0] = 0;
  scanf("%d", &n);
  for (int t = 0; t < n; ++t) {
    scanf("%s", s + 1);
    int cnt = 0, m = strlen(s + 1);
    for (int j = 1; j <= m; ++j) {
      if (s[j] == '(')
        ++cnt;
      else
        --cnt;
      if (cnt <= mn[t]) g[t][-cnt]++, mn[t] = cnt;
    }
    w[t] = cnt;
  }
  for (int s = 0; s < (1 << n); ++s) {
    if (f[s] < 0) continue;
    for (int i = 0; i < n; ++i) {
      if (s >> i & 1) continue;
      int t = s ^ (1 << i);
      ans = max(ans, f[s] + g[i][W[s]]);
      if (W[s] + mn[i] >= 0) {
        W[t] = W[s] + w[i];
        f[t] = max(f[t], f[s] + g[i][W[s]]);
      }
    }
  }
  printf("%d\n", ans);
}

### Prompt

In cpp, your task is to solve 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;
const int MAXN = 4e5 + 7;
char s[MAXN];
int g[20][MAXN], w[20], mn[20], f[1 << 20], W[1 << 20], ans;
int main() {
  memset(f, -0x3f, sizeof(f));
  f[0] = 0;
  scanf("%d", &n);
  for (int t = 0; t < n; ++t) {
    scanf("%s", s + 1);
    int cnt = 0, m = strlen(s + 1);
    for (int j = 1; j <= m; ++j) {
      if (s[j] == '(')
        ++cnt;
      else
        --cnt;
      if (cnt <= mn[t]) g[t][-cnt]++, mn[t] = cnt;
    }
    w[t] = cnt;
  }
  for (int s = 0; s < (1 << n); ++s) {
    if (f[s] < 0) continue;
    for (int i = 0; i < n; ++i) {
      if (s >> i & 1) continue;
      int t = s ^ (1 << i);
      ans = max(ans, f[s] + g[i][W[s]]);
      if (W[s] + mn[i] >= 0) {
        W[t] = W[s] + w[i];
        f[t] = max(f[t], f[s] + g[i][W[s]]);
      }
    }
  }
  printf("%d\n", ans);
}

```