#include <bits/stdc++.h>
using namespace std;
const int M = 1000000007;
const int MM = 998244353;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int _runtimeTerror_() {
  int n;
  cin >> n;
  vector<string> a(n);
  vector<int> len(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    len[i] = (long long)a[i].size();
  }
  vector<map<int, int>> mp(n);
  vector<int> tsum(n, 0);
  for (int i = 0; i < n; ++i) {
    int mn = 0;
    for (char c : a[i]) {
      tsum[i] += c == '(';
      tsum[i] -= c == ')';
      if (tsum[i] <= mn) {
        ++mp[i][tsum[i]];
        amin(mn, tsum[i]);
      }
    }
  }
  vector<long long> dp(1 << n, -2000000000000000000);
  dp[0] = 0;
  vector<long long> sum(1 << n, 0);
  for (int i = 1; i < (1 << n); ++i) {
    for (int j = 0; j < n; ++j) {
      if (i & (1 << j)) {
        sum[i] += tsum[j];
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < (1 << n); ++i) {
    amax(ans, dp[i]);
    if (dp[i] == -2000000000000000000) {
      continue;
    }
    for (int j = 0; j < n; ++j) {
      if ((i & (1 << j)) == 0) {
        int f = mp[j].empty() ? 0 : mp[j].begin()->first;
        if (f + sum[i] < 0) {
          int val = mp[j].count(-sum[i]) ? mp[j][-sum[i]] : 0;
          amax(ans, dp[i] + val);
        } else {
          int val = mp[j].count(-sum[i]) ? mp[j][-sum[i]] : 0;
          amax(dp[i | (1 << j)], dp[i] + val);
        }
      }
    }
  }
  cout << ans << "\n";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 M = 1000000007;
const int MM = 998244353;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
int _runtimeTerror_() {
  int n;
  cin >> n;
  vector<string> a(n);
  vector<int> len(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    len[i] = (long long)a[i].size();
  }
  vector<map<int, int>> mp(n);
  vector<int> tsum(n, 0);
  for (int i = 0; i < n; ++i) {
    int mn = 0;
    for (char c : a[i]) {
      tsum[i] += c == '(';
      tsum[i] -= c == ')';
      if (tsum[i] <= mn) {
        ++mp[i][tsum[i]];
        amin(mn, tsum[i]);
      }
    }
  }
  vector<long long> dp(1 << n, -2000000000000000000);
  dp[0] = 0;
  vector<long long> sum(1 << n, 0);
  for (int i = 1; i < (1 << n); ++i) {
    for (int j = 0; j < n; ++j) {
      if (i & (1 << j)) {
        sum[i] += tsum[j];
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < (1 << n); ++i) {
    amax(ans, dp[i]);
    if (dp[i] == -2000000000000000000) {
      continue;
    }
    for (int j = 0; j < n; ++j) {
      if ((i & (1 << j)) == 0) {
        int f = mp[j].empty() ? 0 : mp[j].begin()->first;
        if (f + sum[i] < 0) {
          int val = mp[j].count(-sum[i]) ? mp[j][-sum[i]] : 0;
          amax(ans, dp[i] + val);
        } else {
          int val = mp[j].count(-sum[i]) ? mp[j][-sum[i]] : 0;
          amax(dp[i | (1 << j)], dp[i] + val);
        }
      }
    }
  }
  cout << ans << "\n";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) _runtimeTerror_();
  return 0;
}

```