#include <bits/stdc++.h>
using namespace std;
string S[20];
vector<pair<pair<int, int>, int> > psList[20];
int ps[(1 << 20)], minps[(1 << 20)], dp[(1 << 20)];
pair<int, int> Process(string &s, vector<pair<pair<int, int>, int> > &psList) {
  int ps = 0, minps = INT_MAX;
  for (int i = 0, n = s.size(); i < n; ++i) {
    ps += (s[i] == '(' ? 1 : -1);
    minps = min(minps, ps);
    psList.push_back({{ps, i}, i});
  }
  sort(psList.begin(), psList.end());
  int x = INT_MAX;
  for (auto &p : psList) {
    x = min(x, p.second);
    p.second = x;
  }
  return {ps, minps};
}
int Calculate(int i, int ps, int c) {
  ps = -ps;
  auto &v = psList[i];
  int k = lower_bound(v.begin(), v.end(), make_pair(make_pair(ps, 0), 0)) -
          v.begin();
  if (k >= v.size() || v[k].first.first != ps) {
    return c;
  }
  int n = k != 0 ? v[k - 1].second : INT_MAX;
  return c +
         (lower_bound(v.begin() + k, v.end(), make_pair(make_pair(ps, n), n)) -
          (v.begin() + k));
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    cin >> S[i];
  }
  for (int i = 0; i < n; ++i) {
    int mask = (1 << i);
    tie(ps[mask], minps[mask]) = Process(S[i], psList[i]);
  }
  int m = (1 << n);
  ps[0] = 0, minps[0] = INT_MAX;
  for (int mask = 1; mask < m; ++mask) {
    int i = __builtin_ctz(mask);
    int imask = (1 << i);
    ps[mask] = ps[(mask ^ imask)] + ps[imask];
  }
  int ans = 0;
  for (int mask = 1; mask < m; ++mask) {
    int ansx = 0, minpsx = -1e6;
    for (int tmask = mask, i = 0; tmask != 0; tmask ^= (1 << i)) {
      i = __builtin_ctz(tmask);
      int rmask = mask ^ (1 << i);
      if (minps[rmask] < 0) {
        continue;
      }
      int ansy = Calculate(i, ps[rmask], dp[rmask]);
      ans = max(ans, ansy);
      int minpsy = min(minps[rmask], ps[rmask] + minps[(mask ^ rmask)]);
      minpsx = max(minpsx, minpsy);
      if (minpsy >= 0) {
        ansx = max(ansx, ansy);
      }
    }
    dp[mask] = ansx;
    minps[mask] = minpsx;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
string S[20];
vector<pair<pair<int, int>, int> > psList[20];
int ps[(1 << 20)], minps[(1 << 20)], dp[(1 << 20)];
pair<int, int> Process(string &s, vector<pair<pair<int, int>, int> > &psList) {
  int ps = 0, minps = INT_MAX;
  for (int i = 0, n = s.size(); i < n; ++i) {
    ps += (s[i] == '(' ? 1 : -1);
    minps = min(minps, ps);
    psList.push_back({{ps, i}, i});
  }
  sort(psList.begin(), psList.end());
  int x = INT_MAX;
  for (auto &p : psList) {
    x = min(x, p.second);
    p.second = x;
  }
  return {ps, minps};
}
int Calculate(int i, int ps, int c) {
  ps = -ps;
  auto &v = psList[i];
  int k = lower_bound(v.begin(), v.end(), make_pair(make_pair(ps, 0), 0)) -
          v.begin();
  if (k >= v.size() || v[k].first.first != ps) {
    return c;
  }
  int n = k != 0 ? v[k - 1].second : INT_MAX;
  return c +
         (lower_bound(v.begin() + k, v.end(), make_pair(make_pair(ps, n), n)) -
          (v.begin() + k));
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    cin >> S[i];
  }
  for (int i = 0; i < n; ++i) {
    int mask = (1 << i);
    tie(ps[mask], minps[mask]) = Process(S[i], psList[i]);
  }
  int m = (1 << n);
  ps[0] = 0, minps[0] = INT_MAX;
  for (int mask = 1; mask < m; ++mask) {
    int i = __builtin_ctz(mask);
    int imask = (1 << i);
    ps[mask] = ps[(mask ^ imask)] + ps[imask];
  }
  int ans = 0;
  for (int mask = 1; mask < m; ++mask) {
    int ansx = 0, minpsx = -1e6;
    for (int tmask = mask, i = 0; tmask != 0; tmask ^= (1 << i)) {
      i = __builtin_ctz(tmask);
      int rmask = mask ^ (1 << i);
      if (minps[rmask] < 0) {
        continue;
      }
      int ansy = Calculate(i, ps[rmask], dp[rmask]);
      ans = max(ans, ansy);
      int minpsy = min(minps[rmask], ps[rmask] + minps[(mask ^ rmask)]);
      minpsx = max(minpsx, minpsy);
      if (minpsy >= 0) {
        ansx = max(ansx, ansy);
      }
    }
    dp[mask] = ansx;
    minps[mask] = minpsx;
  }
  cout << ans << '\n';
  return 0;
}

```