#include <bits/stdc++.h>
using namespace std;
const int N = 20, M = 400000;
int n;
vector<int> cnt[N + 9];
int sum[(1 << N) + 9], mn[N + 9];
void into() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    static char s[M + 9];
    scanf("%s", s);
    int m = strlen(s);
    cnt[i].resize(m + 1);
    for (int j = 0, &pre = sum[1 << i]; j < m; ++j) {
      mn[i] = min(mn[i], pre += s[j] == '(' ? 1 : -1);
      if (pre == mn[i]) ++cnt[i][-pre];
    }
  }
}
int dp[(1 << N) + 9][2];
void Get_dp() {
  for (int s = 1; s < 1 << n; ++s) sum[s] = sum[s & -s] + sum[s ^ s & -s];
  for (int s = 1; s < 1 << n; ++s) {
    dp[s][0] = dp[s][1] = -M;
    for (int i = 0; i < n; ++i) {
      if (s >> i & 1 ^ 1) continue;
      int t = s ^ 1 << i;
      if (sum[t] >= 0 && dp[t][1] != -M) {
        for (int j = sum[t] + mn[i] < 0; j < 2; ++j)
          dp[s][j] =
              max(dp[s][j],
                  dp[t][0] + (sum[t] < cnt[i].size() ? cnt[i][sum[t]] : 0));
      }
      dp[s][1] = max(dp[s][1], dp[t][1]);
    }
  }
}
void work() { Get_dp(); }
void outo() { printf("%d\n", max(dp[(1 << n) - 1][0], dp[(1 << n) - 1][1])); }
int main() {
  int T = 1;
  for (; T--;) {
    into();
    work();
    outo();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int N = 20, M = 400000;
int n;
vector<int> cnt[N + 9];
int sum[(1 << N) + 9], mn[N + 9];
void into() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    static char s[M + 9];
    scanf("%s", s);
    int m = strlen(s);
    cnt[i].resize(m + 1);
    for (int j = 0, &pre = sum[1 << i]; j < m; ++j) {
      mn[i] = min(mn[i], pre += s[j] == '(' ? 1 : -1);
      if (pre == mn[i]) ++cnt[i][-pre];
    }
  }
}
int dp[(1 << N) + 9][2];
void Get_dp() {
  for (int s = 1; s < 1 << n; ++s) sum[s] = sum[s & -s] + sum[s ^ s & -s];
  for (int s = 1; s < 1 << n; ++s) {
    dp[s][0] = dp[s][1] = -M;
    for (int i = 0; i < n; ++i) {
      if (s >> i & 1 ^ 1) continue;
      int t = s ^ 1 << i;
      if (sum[t] >= 0 && dp[t][1] != -M) {
        for (int j = sum[t] + mn[i] < 0; j < 2; ++j)
          dp[s][j] =
              max(dp[s][j],
                  dp[t][0] + (sum[t] < cnt[i].size() ? cnt[i][sum[t]] : 0));
      }
      dp[s][1] = max(dp[s][1], dp[t][1]);
    }
  }
}
void work() { Get_dp(); }
void outo() { printf("%d\n", max(dp[(1 << n) - 1][0], dp[(1 << n) - 1][1])); }
int main() {
  int T = 1;
  for (; T--;) {
    into();
    work();
    outo();
  }
  return 0;
}

```