#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
unordered_map<int, int> cnt[20];
unordered_map<int, int> pre[20];
int sum[1 << 20];
int mi[20];
bool bad[1 << 20];
int dp[1 << 20];
inline int lowbit(int x) { return x & -x; }
void solve() {
  int n;
  cin >> n;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    string str;
    cin >> str;
    int t = 0;
    for (int j = 0, len = str.size(); j < len; j++) {
      t += (str[j] == '(' ? 1 : -1);
      mi[i] = min(mi[i], t);
      cnt[i][t]++;
      if (pre[i].find(t) == pre[i].end()) pre[i][t] = cnt[i][t + 1];
    }
    sum[1 << i] = t;
  }
  for (int i = 1; i < (1 << n); i++) {
    sum[i] = sum[i ^ lowbit(i)] + sum[lowbit(i)];
    bool ok = false;
    for (int j = 0; j < n; j++)
      if ((i >> j) & 1) {
        int k = (1 << j);
        if (dp[i ^ k] < 0) continue;
        int d = sum[i ^ k];
        if (d + mi[j] < 0) {
          ans = max(ans, dp[i ^ k] + pre[j][-d - 1]);
        } else {
          dp[i] = max(dp[i], dp[i ^ k] + cnt[j][-d]);
          ok = true;
        }
      }
    ans = max(ans, dp[i]);
    if (!ok) dp[i] = -1;
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  solve();
  return 0;
}

### 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;
const int INF = 0x3f3f3f3f;
unordered_map<int, int> cnt[20];
unordered_map<int, int> pre[20];
int sum[1 << 20];
int mi[20];
bool bad[1 << 20];
int dp[1 << 20];
inline int lowbit(int x) { return x & -x; }
void solve() {
  int n;
  cin >> n;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    string str;
    cin >> str;
    int t = 0;
    for (int j = 0, len = str.size(); j < len; j++) {
      t += (str[j] == '(' ? 1 : -1);
      mi[i] = min(mi[i], t);
      cnt[i][t]++;
      if (pre[i].find(t) == pre[i].end()) pre[i][t] = cnt[i][t + 1];
    }
    sum[1 << i] = t;
  }
  for (int i = 1; i < (1 << n); i++) {
    sum[i] = sum[i ^ lowbit(i)] + sum[lowbit(i)];
    bool ok = false;
    for (int j = 0; j < n; j++)
      if ((i >> j) & 1) {
        int k = (1 << j);
        if (dp[i ^ k] < 0) continue;
        int d = sum[i ^ k];
        if (d + mi[j] < 0) {
          ans = max(ans, dp[i ^ k] + pre[j][-d - 1]);
        } else {
          dp[i] = max(dp[i], dp[i ^ k] + cnt[j][-d]);
          ok = true;
        }
      }
    ans = max(ans, dp[i]);
    if (!ok) dp[i] = -1;
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  solve();
  return 0;
}

```