#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return l + rng() % (r - l + 1); }
const int N = 21, TT = 1 << 20;
int n;
int f[TT], num[TT], l[N];
string a[N];
vector<int> s[N], m[N], c[N];
int main() {
  if (fopen("TASK"
            ".inp",
            "r")) {
    freopen(
        "TASK"
        ".inp",
        "r", stdin);
  }
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = 1, _c = n; i <= _c; ++i) {
    cin >> a[i];
    l[i] = a[i].size();
    a[i] = ' ' + a[i];
    s[i].resize(l[i] + 1);
    m[i].resize(l[i] + 1);
    c[i].resize(l[i] + 1);
    m[i][0] = 1e9;
    for (int j = 1, _c = l[i]; j <= _c; ++j) {
      s[i][j] = s[i][j - 1] + ((a[i][j] == '(') ? 1 : -1);
      m[i][j] = m[i][j - 1];
      c[i][j] = c[i][j - 1];
      if (s[i][j] < m[i][j - 1]) {
        m[i][j] = s[i][j];
        c[i][j] = 0;
      }
      if (s[i][j] == m[i][j]) c[i][j]++;
    }
  }
  memset(f, -1, sizeof f);
  f[0] = 0;
  int oo = (1 << n) - 1, ans = 0;
  for (int t = 0, _c = oo - 1; t <= _c; ++t)
    if (f[t] != -1) {
      for (int i = 1, _c = n; i <= _c; ++i)
        if (!((t >> (i - 1)) & 1)) {
          int t2 = (t | (1ll << (i - 1)));
          num[t2] = num[t] + s[i][l[i]];
          if (num[t] + m[i][l[i]] >= 0) {
            int sl = 0;
            if (num[t] + m[i][l[i]] == 0) sl += c[i][l[i]];
            f[t2] = max(f[t2], f[t] + sl);
            ans = max(ans, f[t2]);
          } else {
            int lo = 1, hi = l[i], ret = 0;
            while (lo <= hi) {
              int mid = (lo + hi) / 2;
              if (num[t] + m[i][mid] >= 0)
                ret = mid, lo = mid + 1;
              else
                hi = mid - 1;
            }
            ans = max(ans, f[t] + c[i][ret]);
          }
        }
    }
  cout << ans << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return l + rng() % (r - l + 1); }
const int N = 21, TT = 1 << 20;
int n;
int f[TT], num[TT], l[N];
string a[N];
vector<int> s[N], m[N], c[N];
int main() {
  if (fopen("TASK"
            ".inp",
            "r")) {
    freopen(
        "TASK"
        ".inp",
        "r", stdin);
  }
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = 1, _c = n; i <= _c; ++i) {
    cin >> a[i];
    l[i] = a[i].size();
    a[i] = ' ' + a[i];
    s[i].resize(l[i] + 1);
    m[i].resize(l[i] + 1);
    c[i].resize(l[i] + 1);
    m[i][0] = 1e9;
    for (int j = 1, _c = l[i]; j <= _c; ++j) {
      s[i][j] = s[i][j - 1] + ((a[i][j] == '(') ? 1 : -1);
      m[i][j] = m[i][j - 1];
      c[i][j] = c[i][j - 1];
      if (s[i][j] < m[i][j - 1]) {
        m[i][j] = s[i][j];
        c[i][j] = 0;
      }
      if (s[i][j] == m[i][j]) c[i][j]++;
    }
  }
  memset(f, -1, sizeof f);
  f[0] = 0;
  int oo = (1 << n) - 1, ans = 0;
  for (int t = 0, _c = oo - 1; t <= _c; ++t)
    if (f[t] != -1) {
      for (int i = 1, _c = n; i <= _c; ++i)
        if (!((t >> (i - 1)) & 1)) {
          int t2 = (t | (1ll << (i - 1)));
          num[t2] = num[t] + s[i][l[i]];
          if (num[t] + m[i][l[i]] >= 0) {
            int sl = 0;
            if (num[t] + m[i][l[i]] == 0) sl += c[i][l[i]];
            f[t2] = max(f[t2], f[t] + sl);
            ans = max(ans, f[t2]);
          } else {
            int lo = 1, hi = l[i], ret = 0;
            while (lo <= hi) {
              int mid = (lo + hi) / 2;
              if (num[t] + m[i][mid] >= 0)
                ret = mid, lo = mid + 1;
              else
                hi = mid - 1;
            }
            ans = max(ans, f[t] + c[i][ret]);
          }
        }
    }
  cout << ans << "\n";
  return 0;
}

```