#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<string> s(n);
  vector<int> bal(n);
  vector<int> mn(n);
  vector<unordered_map<int, int>> get(n);
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    int sum = 0;
    int mnn = 0;
    for (int j = 0; j < s[i].size(); ++j) {
      if (s[i][j] == '(') {
        ++sum;
      } else {
        --sum;
      }
      mnn = min(mnn, sum);
      if (mnn >= sum) {
        get[i][sum] += 1;
      }
    }
    bal[i] = sum;
    mn[i] = mnn;
  }
  vector<vector<int>> v(n, vector<int>(1 << n));
  vector<int> sum(1 << n);
  for (int mask = 0; mask < (1 << n); ++mask) {
    for (int j = 0; j < n; ++j) {
      if ((mask >> j) & 1) {
        sum[mask] = sum[mask ^ (1 << j)] + bal[j];
        break;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int mask = 0; mask < (1 << n); ++mask) {
      int sz = sum[mask];
      if ((mask >> i) & 1) {
        sz -= bal[i];
        continue;
      }
      v[i][mask] = get[i][-sz];
    }
  }
  vector<vector<int>> dp(1 << n, vector<int>(2, -1e9));
  dp[0][0] = 0;
  for (int mask = 0; mask < (1 << n); ++mask) {
    for (int j = 0; j < n; ++j) {
      if ((mask >> j) & 1) {
        if (sum[mask ^ (1 << j)] + mn[j] >= 0) {
          dp[mask][0] =
              max(dp[mask][0], dp[mask ^ (1 << j)][0] + v[j][mask ^ (1 << j)]);
        } else {
          dp[mask][1] =
              max(dp[mask][1], dp[mask ^ (1 << j)][0] + v[j][mask ^ (1 << j)]);
        }
        dp[mask][1] = max(dp[mask][1], dp[mask ^ (1 << j)][1]);
      }
    }
  }
  cout << *max_element((dp[(1 << n) - 1]).begin(), (dp[(1 << n) - 1]).end());
}

### 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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<string> s(n);
  vector<int> bal(n);
  vector<int> mn(n);
  vector<unordered_map<int, int>> get(n);
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    int sum = 0;
    int mnn = 0;
    for (int j = 0; j < s[i].size(); ++j) {
      if (s[i][j] == '(') {
        ++sum;
      } else {
        --sum;
      }
      mnn = min(mnn, sum);
      if (mnn >= sum) {
        get[i][sum] += 1;
      }
    }
    bal[i] = sum;
    mn[i] = mnn;
  }
  vector<vector<int>> v(n, vector<int>(1 << n));
  vector<int> sum(1 << n);
  for (int mask = 0; mask < (1 << n); ++mask) {
    for (int j = 0; j < n; ++j) {
      if ((mask >> j) & 1) {
        sum[mask] = sum[mask ^ (1 << j)] + bal[j];
        break;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int mask = 0; mask < (1 << n); ++mask) {
      int sz = sum[mask];
      if ((mask >> i) & 1) {
        sz -= bal[i];
        continue;
      }
      v[i][mask] = get[i][-sz];
    }
  }
  vector<vector<int>> dp(1 << n, vector<int>(2, -1e9));
  dp[0][0] = 0;
  for (int mask = 0; mask < (1 << n); ++mask) {
    for (int j = 0; j < n; ++j) {
      if ((mask >> j) & 1) {
        if (sum[mask ^ (1 << j)] + mn[j] >= 0) {
          dp[mask][0] =
              max(dp[mask][0], dp[mask ^ (1 << j)][0] + v[j][mask ^ (1 << j)]);
        } else {
          dp[mask][1] =
              max(dp[mask][1], dp[mask ^ (1 << j)][0] + v[j][mask ^ (1 << j)]);
        }
        dp[mask][1] = max(dp[mask][1], dp[mask ^ (1 << j)][1]);
      }
    }
  }
  cout << *max_element((dp[(1 << n) - 1]).begin(), (dp[(1 << n) - 1]).end());
}

```