#include <bits/stdc++.h>
using namespace std;
int n, dp[1 << 20][2], S[1 << 20], cnt[20 + 1], length[20 + 1], lg[1 << 20];
string s[20 + 1];
vector<int> minn[20 + 1];
vector<vector<int>> p[20 + 1];
int lowbit(int x) { return x & (-x); }
int F(int x, int d) {
  if (d < -length[x] || d > length[x]) return 0;
  int ps =
      lower_bound(minn[x].begin(), minn[x].end(), d + 1) - minn[x].begin() - 1;
  return lower_bound(p[x][length[x] + d].begin(), p[x][length[x] + d].end(),
                     ps + 1) -
         p[x][length[x] + d].begin();
}
int main() {
  cin >> n, dp[0][1] = -1e9;
  for (int i = 1; i < (1 << n); ++i) dp[i][0] = dp[i][1] = -1e9;
  for (int i = 2; i < (1 << n); ++i) lg[i] = lg[i >> 1] + 1;
  for (int i = 1; i <= n; ++i) {
    cin >> s[i], length[i] = s[i].length(), minn[i].resize(length[i] + 1),
                 p[i].resize(2 * length[i] + 1);
    for (int j = 1; j <= length[i]; ++j) {
      if (s[i][j - 1] == '(')
        cnt[i]++;
      else
        cnt[i]--;
      minn[i][j] = min(minn[i][j - 1], cnt[i]),
      p[i][length[i] - cnt[i]].push_back(j);
    }
    for (int j = 0; j <= length[i]; ++j) minn[i][j] = -minn[i][j];
  }
  for (int i = 1; i < (1 << n); ++i) {
    S[i] = S[i - lowbit(i)] + cnt[lg[lowbit(i)] + 1];
    for (int j = 1; j <= n; ++j)
      if (i & (1 << (j - 1))) {
        bool op = (S[i - (1 << (j - 1))] - minn[j][length[j]] < 0);
        dp[i][op] = max(
            dp[i][op], dp[i - (1 << (j - 1))][0] + F(j, S[i - (1 << (j - 1))]));
        dp[i][1] = max(dp[i][1], dp[i - (1 << (j - 1))][1]);
      }
  }
  cout << max(dp[(1 << n) - 1][0], dp[(1 << n) - 1][1]) << endl;
  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;
int n, dp[1 << 20][2], S[1 << 20], cnt[20 + 1], length[20 + 1], lg[1 << 20];
string s[20 + 1];
vector<int> minn[20 + 1];
vector<vector<int>> p[20 + 1];
int lowbit(int x) { return x & (-x); }
int F(int x, int d) {
  if (d < -length[x] || d > length[x]) return 0;
  int ps =
      lower_bound(minn[x].begin(), minn[x].end(), d + 1) - minn[x].begin() - 1;
  return lower_bound(p[x][length[x] + d].begin(), p[x][length[x] + d].end(),
                     ps + 1) -
         p[x][length[x] + d].begin();
}
int main() {
  cin >> n, dp[0][1] = -1e9;
  for (int i = 1; i < (1 << n); ++i) dp[i][0] = dp[i][1] = -1e9;
  for (int i = 2; i < (1 << n); ++i) lg[i] = lg[i >> 1] + 1;
  for (int i = 1; i <= n; ++i) {
    cin >> s[i], length[i] = s[i].length(), minn[i].resize(length[i] + 1),
                 p[i].resize(2 * length[i] + 1);
    for (int j = 1; j <= length[i]; ++j) {
      if (s[i][j - 1] == '(')
        cnt[i]++;
      else
        cnt[i]--;
      minn[i][j] = min(minn[i][j - 1], cnt[i]),
      p[i][length[i] - cnt[i]].push_back(j);
    }
    for (int j = 0; j <= length[i]; ++j) minn[i][j] = -minn[i][j];
  }
  for (int i = 1; i < (1 << n); ++i) {
    S[i] = S[i - lowbit(i)] + cnt[lg[lowbit(i)] + 1];
    for (int j = 1; j <= n; ++j)
      if (i & (1 << (j - 1))) {
        bool op = (S[i - (1 << (j - 1))] - minn[j][length[j]] < 0);
        dp[i][op] = max(
            dp[i][op], dp[i - (1 << (j - 1))][0] + F(j, S[i - (1 << (j - 1))]));
        dp[i][1] = max(dp[i][1], dp[i - (1 << (j - 1))][1]);
      }
  }
  cout << max(dp[(1 << n) - 1][0], dp[(1 << n) - 1][1]) << endl;
  return 0;
}

```