#include <bits/stdc++.h>
using namespace std;
int n;
string s[21];
int mp[21][400005];
int baki[21];
int mx[21];
int dp[1 << 21];
int run(int mask) {
  if (mask == (1 << n) - 1) {
    return 0;
  }
  if (dp[mask] != -1) return dp[mask];
  int a = 0;
  for (int i = 0; i < n; i++) {
    if (mask & (1 << i)) {
      a += baki[i] - mx[i];
    }
  }
  int tot = 0;
  for (int i = 0; i < n; i++) {
    if (mask & (1 << i)) continue;
    int ans = 0;
    ans += mp[i][a];
    if (mx[i] <= a) {
      ans += run(mask | (1 << i));
    }
    tot = max(ans, tot);
  }
  return dp[mask] = tot;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    int a = 0, b = 0;
    int ex = 0;
    for (auto u : s[i]) {
      if (u == '(') {
        a++;
      } else {
        if (a == 0) {
          ex++;
          mp[i][ex]++;
        } else {
          a--;
          if (a == 0) mp[i][ex]++;
        }
      }
      mx[i] = max(mx[i], ex);
    }
    baki[i] = a;
  }
  memset(dp, -1, sizeof dp);
  cout << run(0);
}

### Prompt

Please formulate 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;
int n;
string s[21];
int mp[21][400005];
int baki[21];
int mx[21];
int dp[1 << 21];
int run(int mask) {
  if (mask == (1 << n) - 1) {
    return 0;
  }
  if (dp[mask] != -1) return dp[mask];
  int a = 0;
  for (int i = 0; i < n; i++) {
    if (mask & (1 << i)) {
      a += baki[i] - mx[i];
    }
  }
  int tot = 0;
  for (int i = 0; i < n; i++) {
    if (mask & (1 << i)) continue;
    int ans = 0;
    ans += mp[i][a];
    if (mx[i] <= a) {
      ans += run(mask | (1 << i));
    }
    tot = max(ans, tot);
  }
  return dp[mask] = tot;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    int a = 0, b = 0;
    int ex = 0;
    for (auto u : s[i]) {
      if (u == '(') {
        a++;
      } else {
        if (a == 0) {
          ex++;
          mp[i][ex]++;
        } else {
          a--;
          if (a == 0) mp[i][ex]++;
        }
      }
      mx[i] = max(mx[i], ex);
    }
    baki[i] = a;
  }
  memset(dp, -1, sizeof dp);
  cout << run(0);
}

```