#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 22, MAXM = 4e5 + 10, INF = 1e16;
int n, len[MAXN], sum[MAXN][MAXM], pre[MAXN][MAXM];
long long f[(1 << 20) + 5], ans;
char s[MAXN][MAXM];
vector<int> occ[MAXN][MAXM];
void update(int i, int s, int j) {
  int L = 1, R = len[j], pos = 1e9, ret = 0;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (pre[j][mid] < -s) {
      pos = mid;
      R = mid - 1;
    } else {
      L = mid + 1;
    }
  }
  ret =
      lower_bound(occ[j][s].begin(), occ[j][s].end(), pos) - occ[j][s].begin();
  ans = max(ans, f[i] + ret);
  if (pos == 1e9)
    f[i | (1 << (j - 1))] = max(f[i | (1 << (j - 1))], f[i] + ret);
}
void dp() {
  for (int i = (0); i <= ((1 << n) - 1); i++) {
    int s = 0;
    for (int j = (0); j <= (n - 1); j++)
      if ((i >> j) & 1) s += sum[j + 1][len[j + 1]];
    if (s < 0) continue;
    for (int j = (0); j <= (n - 1); j++)
      if (!((i >> j) & 1)) update(i, s, j + 1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) {
    scanf("%s", s[i] + 1);
    len[i] = strlen(s[i] + 1);
  }
  for (int i = (1); i <= (n); i++) {
    pre[i][0] = 1e9;
    for (int j = (1); j <= (len[i]); j++) {
      sum[i][j] = sum[i][j - 1];
      if (s[i][j] == '(')
        sum[i][j]++;
      else
        sum[i][j]--;
      pre[i][j] = min(pre[i][j - 1], sum[i][j]);
    }
    for (int j = (1); j <= (len[i]); j++) {
      if (sum[i][j] <= 0) occ[i][-sum[i][j]].push_back(j);
    }
  }
  for (int i = (1); i <= ((1 << n) - 1); i++) f[i] = -INF;
  ans = 0;
  dp();
  for (int i = (1); i <= ((1 << n) - 1); i++) ans = max(ans, f[i]);
  printf("%lld", ans);
  return 0;
}

### Prompt

Generate 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;
const long long MAXN = 22, MAXM = 4e5 + 10, INF = 1e16;
int n, len[MAXN], sum[MAXN][MAXM], pre[MAXN][MAXM];
long long f[(1 << 20) + 5], ans;
char s[MAXN][MAXM];
vector<int> occ[MAXN][MAXM];
void update(int i, int s, int j) {
  int L = 1, R = len[j], pos = 1e9, ret = 0;
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (pre[j][mid] < -s) {
      pos = mid;
      R = mid - 1;
    } else {
      L = mid + 1;
    }
  }
  ret =
      lower_bound(occ[j][s].begin(), occ[j][s].end(), pos) - occ[j][s].begin();
  ans = max(ans, f[i] + ret);
  if (pos == 1e9)
    f[i | (1 << (j - 1))] = max(f[i | (1 << (j - 1))], f[i] + ret);
}
void dp() {
  for (int i = (0); i <= ((1 << n) - 1); i++) {
    int s = 0;
    for (int j = (0); j <= (n - 1); j++)
      if ((i >> j) & 1) s += sum[j + 1][len[j + 1]];
    if (s < 0) continue;
    for (int j = (0); j <= (n - 1); j++)
      if (!((i >> j) & 1)) update(i, s, j + 1);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) {
    scanf("%s", s[i] + 1);
    len[i] = strlen(s[i] + 1);
  }
  for (int i = (1); i <= (n); i++) {
    pre[i][0] = 1e9;
    for (int j = (1); j <= (len[i]); j++) {
      sum[i][j] = sum[i][j - 1];
      if (s[i][j] == '(')
        sum[i][j]++;
      else
        sum[i][j]--;
      pre[i][j] = min(pre[i][j - 1], sum[i][j]);
    }
    for (int j = (1); j <= (len[i]); j++) {
      if (sum[i][j] <= 0) occ[i][-sum[i][j]].push_back(j);
    }
  }
  for (int i = (1); i <= ((1 << n) - 1); i++) f[i] = -INF;
  ans = 0;
  dp();
  for (int i = (1); i <= ((1 << n) - 1); i++) ans = max(ans, f[i]);
  printf("%lld", ans);
  return 0;
}

```