#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
inline long long llread() {
  long long x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
int n;
char s[25][400010];
int l[25], rest[25];
vector<int> v[25][400010];
int dp[(1 << 20) + 10][2], g[(1 << 20) + 10];
signed main() {
  n = read();
  for (int i = (1); i <= (n); i++) scanf("%s", s[i] + 1);
  for (int i = (1); i <= (n); i++) l[i] = strlen(s[i] + 1);
  for (int i = (1); i <= (n); i++) {
    int top = 0;
    for (int j = (1); j <= (l[i]); j++) {
      if (s[i][j] == '(')
        top++;
      else
        top--;
      if (top <= 0) v[i][-top].push_back(j);
    }
    rest[i] = top;
  }
  for (int i = (1); i <= ((1 << n) - 1); i++) dp[i][0] = dp[i][1] = -0x3f3f3f3f;
  dp[0][0] = dp[0][1] = g[0] = 0;
  for (int mask = (1); mask <= ((1 << n) - 1); mask++) {
    for (int i = (1); i <= (n); i++) {
      if (mask & (1 << (i - 1))) {
        int S = mask ^ (1 << (i - 1));
        if (dp[mask][0] < dp[S][0]) {
          dp[mask][0] = dp[S][0];
        };
        if (g[S] >= 0) {
          int x;
          if (v[i][g[S] + 1].size() == 0)
            x = l[i] + 1;
          else
            x = v[i][g[S] + 1][0];
          int y = lower_bound(v[i][g[S]].begin(), v[i][g[S]].end(), x) -
                  v[i][g[S]].begin();
          if (x == l[i] + 1) {
            if (dp[mask][1] < dp[S][1] + y) {
              dp[mask][1] = dp[S][1] + y;
            };
          }
          if (dp[mask][0] < dp[S][1] + y) {
            dp[mask][0] = dp[S][1] + y;
          };
        }
        g[mask] = g[S] + rest[i];
      }
    }
  }
  printf("%d\n", max(dp[(1 << n) - 1][0], dp[(1 << n) - 1][1]));
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
inline int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
inline long long llread() {
  long long x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
  return f ? -x : x;
}
int n;
char s[25][400010];
int l[25], rest[25];
vector<int> v[25][400010];
int dp[(1 << 20) + 10][2], g[(1 << 20) + 10];
signed main() {
  n = read();
  for (int i = (1); i <= (n); i++) scanf("%s", s[i] + 1);
  for (int i = (1); i <= (n); i++) l[i] = strlen(s[i] + 1);
  for (int i = (1); i <= (n); i++) {
    int top = 0;
    for (int j = (1); j <= (l[i]); j++) {
      if (s[i][j] == '(')
        top++;
      else
        top--;
      if (top <= 0) v[i][-top].push_back(j);
    }
    rest[i] = top;
  }
  for (int i = (1); i <= ((1 << n) - 1); i++) dp[i][0] = dp[i][1] = -0x3f3f3f3f;
  dp[0][0] = dp[0][1] = g[0] = 0;
  for (int mask = (1); mask <= ((1 << n) - 1); mask++) {
    for (int i = (1); i <= (n); i++) {
      if (mask & (1 << (i - 1))) {
        int S = mask ^ (1 << (i - 1));
        if (dp[mask][0] < dp[S][0]) {
          dp[mask][0] = dp[S][0];
        };
        if (g[S] >= 0) {
          int x;
          if (v[i][g[S] + 1].size() == 0)
            x = l[i] + 1;
          else
            x = v[i][g[S] + 1][0];
          int y = lower_bound(v[i][g[S]].begin(), v[i][g[S]].end(), x) -
                  v[i][g[S]].begin();
          if (x == l[i] + 1) {
            if (dp[mask][1] < dp[S][1] + y) {
              dp[mask][1] = dp[S][1] + y;
            };
          }
          if (dp[mask][0] < dp[S][1] + y) {
            dp[mask][0] = dp[S][1] + y;
          };
        }
        g[mask] = g[S] + rest[i];
      }
    }
  }
  printf("%d\n", max(dp[(1 << n) - 1][0], dp[(1 << n) - 1][1]));
  return 0;
}

```