#include <bits/stdc++.h>
using namespace std;
const int MAXN = (1ll << 21);
long long sum[MAXN], mins[22];
vector<vector<long long>> vm;
long long dp[MAXN];
long long f(long long x) {
  long long &res = dp[x];
  if (res >= 0) return res;
  if (x == (1ll << ((int)vm.size())) - 1) return res = 0;
  res = 0;
  long long s = sum[x];
  for (int i = 0, ggdem = ((int)vm.size()); i < ggdem; ++i) {
    if (!((1ll << i) & x)) {
      long long val = 0;
      if (mins[i] + s <= 0) {
        val += vm[i][s];
      }
      if (mins[i] + s >= 0) {
        val += f(x ^ (1ll << i));
      }
      res = max(res, val);
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<string> s(n);
  for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> s[i];
  vm.resize(n);
  for (int i = 0, ggdem = n; i < ggdem; ++i) {
    long long mini = 0, va = 0;
    vm[i] = {0};
    for (auto j : s[i]) {
      if (j == '(')
        va++;
      else
        va--;
      if (va == mini) vm[i].back()++;
      if (va < mini) vm[i].push_back(1);
      mini = min(mini, va);
    }
    mins[i] = mini;
    sum[(1ll << i)] = va;
  }
  for (int i = 1, ggdem = (1ll << n); i < ggdem; ++i) {
    for (int j = 0, ggdem = n; j < ggdem; ++j) {
      if ((1ll << j) & i) {
        sum[i] = sum[i ^ (1ll << j)] + sum[(1ll << j)];
      }
    }
  }
  memset((dp), (-1), sizeof(dp));
  cout << f(0) << "\n";
  return 0;
}

### Prompt

Your task is to create 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;
const int MAXN = (1ll << 21);
long long sum[MAXN], mins[22];
vector<vector<long long>> vm;
long long dp[MAXN];
long long f(long long x) {
  long long &res = dp[x];
  if (res >= 0) return res;
  if (x == (1ll << ((int)vm.size())) - 1) return res = 0;
  res = 0;
  long long s = sum[x];
  for (int i = 0, ggdem = ((int)vm.size()); i < ggdem; ++i) {
    if (!((1ll << i) & x)) {
      long long val = 0;
      if (mins[i] + s <= 0) {
        val += vm[i][s];
      }
      if (mins[i] + s >= 0) {
        val += f(x ^ (1ll << i));
      }
      res = max(res, val);
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<string> s(n);
  for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> s[i];
  vm.resize(n);
  for (int i = 0, ggdem = n; i < ggdem; ++i) {
    long long mini = 0, va = 0;
    vm[i] = {0};
    for (auto j : s[i]) {
      if (j == '(')
        va++;
      else
        va--;
      if (va == mini) vm[i].back()++;
      if (va < mini) vm[i].push_back(1);
      mini = min(mini, va);
    }
    mins[i] = mini;
    sum[(1ll << i)] = va;
  }
  for (int i = 1, ggdem = (1ll << n); i < ggdem; ++i) {
    for (int j = 0, ggdem = n; j < ggdem; ++j) {
      if ((1ll << j) & i) {
        sum[i] = sum[i ^ (1ll << j)] + sum[(1ll << j)];
      }
    }
  }
  memset((dp), (-1), sizeof(dp));
  cout << f(0) << "\n";
  return 0;
}

```