#include <bits/stdc++.h>
using namespace std;
const int maxn = 21, maxv = 400005;
int n, m, ans, tot;
int f[1 << maxn], mn[maxn], cnt[maxn], sum[1 << maxn], w[maxn][maxv << 1],
    ok[maxn][maxv << 1];
string s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int now = 0;
    cin >> s;
    for (int j = 0; j < s.size(); j++) {
      if (s[j] == '(') now++;
      if (s[j] == ')') ok[i][400000 + now] = 1, now--;
      if (ok[i][400000 + now] == 0) w[i][400000 + now]++;
      mn[i] = min(mn[i], now);
    }
    sum[1 << (i - 1)] = now;
  }
  memset(f, -1, sizeof(f)), f[0] = 0;
  for (int i = 1; i < (1 << n); i++) {
    sum[i] = sum[i ^ i & -i] + sum[i & -i];
    for (int j = 1; j <= n; j++)
      if ((i >> (j - 1)) & 1) {
        int pre = (i ^ (1 << (j - 1)));
        if (f[pre] == -1) continue;
        if (sum[pre] + mn[j] >= 0)
          f[i] = max(f[i], f[pre] + w[j][400000 - sum[pre]]);
        else
          ans = max(ans, f[pre] + w[j][400000 - sum[pre]]);
      }
    ans = max(ans, f[i]);
  }
  printf("%d\n", ans);
  return 0;
}

### 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;
const int maxn = 21, maxv = 400005;
int n, m, ans, tot;
int f[1 << maxn], mn[maxn], cnt[maxn], sum[1 << maxn], w[maxn][maxv << 1],
    ok[maxn][maxv << 1];
string s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int now = 0;
    cin >> s;
    for (int j = 0; j < s.size(); j++) {
      if (s[j] == '(') now++;
      if (s[j] == ')') ok[i][400000 + now] = 1, now--;
      if (ok[i][400000 + now] == 0) w[i][400000 + now]++;
      mn[i] = min(mn[i], now);
    }
    sum[1 << (i - 1)] = now;
  }
  memset(f, -1, sizeof(f)), f[0] = 0;
  for (int i = 1; i < (1 << n); i++) {
    sum[i] = sum[i ^ i & -i] + sum[i & -i];
    for (int j = 1; j <= n; j++)
      if ((i >> (j - 1)) & 1) {
        int pre = (i ^ (1 << (j - 1)));
        if (f[pre] == -1) continue;
        if (sum[pre] + mn[j] >= 0)
          f[i] = max(f[i], f[pre] + w[j][400000 - sum[pre]]);
        else
          ans = max(ans, f[pre] + w[j][400000 - sum[pre]]);
      }
    ans = max(ans, f[i]);
  }
  printf("%d\n", ans);
  return 0;
}

```