#include <bits/stdc++.h>
const int N = 500005, inf = 0x3f3f3f3f;
using namespace std;
int n, f[1 << 20], sm[1 << 20], a[25][N], b[25][N], c[25][N], ans, len[25];
char s[25][N];
void init() {
  for (register int i = 1; i <= n; i++) {
    for (register int j = 1; j <= len[i]; j++) {
      a[i][j] = a[i][j - 1] + (s[i][j] == '(' ? 1 : -1);
      b[i][j] = b[i][j - 1], c[i][j] = c[i][j - 1];
      if (a[i][j] < b[i][j])
        b[i][j] = a[i][j], c[i][j] = 1;
      else if (a[i][j] == b[i][j])
        c[i][j]++;
    }
  }
}
int get(int i, int mn) {
  int l = 0, r = len[i], res = 0;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (b[i][mid] >= mn)
      res = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return res;
}
void solve() {
  memset(f, -inf, sizeof f), f[0] = 0;
  for (register int s = 1; s <= (1 << n) - 1; s++) {
    for (register int i = 1; i <= n; i++) {
      if (!((s >> (i - 1)) & 1)) continue;
      int t = s ^ (1 << (i - 1));
      if (f[t] == -inf) continue;
      int mn = sm[t] + b[i][len[i]];
      if (mn >= 0) {
        if (f[s] < f[t] + (mn ? 0 : c[i][len[i]])) {
          f[s] = f[t] + (mn ? 0 : c[i][len[i]]);
          sm[s] = sm[t] + a[i][len[i]];
        }
      }
      mn = -sm[t];
      int p = get(i, mn);
      ans = max(ans, f[t] + (b[i][p] == mn ? c[i][p] : 0));
    }
  }
}
signed main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%s", s[i] + 1), len[i] = strlen(s[i] + 1);
  }
  init();
  solve();
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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>
const int N = 500005, inf = 0x3f3f3f3f;
using namespace std;
int n, f[1 << 20], sm[1 << 20], a[25][N], b[25][N], c[25][N], ans, len[25];
char s[25][N];
void init() {
  for (register int i = 1; i <= n; i++) {
    for (register int j = 1; j <= len[i]; j++) {
      a[i][j] = a[i][j - 1] + (s[i][j] == '(' ? 1 : -1);
      b[i][j] = b[i][j - 1], c[i][j] = c[i][j - 1];
      if (a[i][j] < b[i][j])
        b[i][j] = a[i][j], c[i][j] = 1;
      else if (a[i][j] == b[i][j])
        c[i][j]++;
    }
  }
}
int get(int i, int mn) {
  int l = 0, r = len[i], res = 0;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (b[i][mid] >= mn)
      res = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  return res;
}
void solve() {
  memset(f, -inf, sizeof f), f[0] = 0;
  for (register int s = 1; s <= (1 << n) - 1; s++) {
    for (register int i = 1; i <= n; i++) {
      if (!((s >> (i - 1)) & 1)) continue;
      int t = s ^ (1 << (i - 1));
      if (f[t] == -inf) continue;
      int mn = sm[t] + b[i][len[i]];
      if (mn >= 0) {
        if (f[s] < f[t] + (mn ? 0 : c[i][len[i]])) {
          f[s] = f[t] + (mn ? 0 : c[i][len[i]]);
          sm[s] = sm[t] + a[i][len[i]];
        }
      }
      mn = -sm[t];
      int p = get(i, mn);
      ans = max(ans, f[t] + (b[i][p] == mn ? c[i][p] : 0));
    }
  }
}
signed main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) {
    scanf("%s", s[i] + 1), len[i] = strlen(s[i] + 1);
  }
  init();
  solve();
  printf("%d\n", ans);
  return 0;
}

```