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.
1
)(((((()((())(((((((
 1
)()())((()))((()((()
 10
((((((((((((((((()))))))))))))))))
(((((((((((((((())))))))))))))))
((((((((((((((((((((()))))))))))))))))))))
(((((((((((((((((((((())))))))))))))))))))))
()()()()()()()()()()()()()()()()()()()()
()()()()()()()()()()()()()()()()()()()()()()()()()()
(((((((((((((((((())))))))))))))))))
()()()()()()()()()()()()()()()()()()()
()()()()()()()()()()()()()()()()()()
()()()()()()()()()()()()()()()()()()()()()()()
 20
)
(
)
)
)
)
)
(
(
)
)
)
)
(
)
)
)
)
(
)
 6
()()()()()
()()()()
()
((((()))))
((()))
()()
 1
((((((((((((((((((((
 10
((((((((((((((((((((((((()))))))))))))))))))))))))
()()()()()()()()()()()()()()()()()()()
(((((((((((((((((())))))))))))))))))
()()()()()()()()()()()()()()()()()()()()()()
((((((((((((((((((()))))))))))))))))))
((((((((((((((((((((((()))))))))))))))))))))))
()()()()()()()()()()()()()()()()()()()
((((((((((((((()))))))))))))))
()()()()()()()()()()()()()()()()()()()()()()()()
()()()()()()()()()()()()()()()()
 2
))())))((()
()((((((()))))))(()
 20
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
)
 20
(
(
)
(
(
)
(
(
(
(
(
(
)
)
(
(
(
(
(
)
 10
()()
(())
()()
()()
(())
(((())))
()
(())
()()
()
 1
))))))))))))))))))))
 6
(((())))
(((())))
()()()
(())
()()()()()
()()
 10
)
)
))
))
))
))
))
))
)))
)))
 10
)()
()
)
)
)
)(
)))
)(
(((
))
 20
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
 10
))
()(
(
()
()
((((
(
(
(((
(
 10
()
()()()()
(())
((()))
(())
()()
(())
()()
()
()
 10
())
()
)
)))
))
()
)
))
)))
)
 1
))))))()))())(()))))
 10
((
(((
(
(((
((((
((
(
(
(
((
 20
)
(
(
)
)
(
)
(
(
(
)
)
)
)
(
)
(
(
)
(

0
 0
 111
 5
 14
 0
 105
 3
 0
 5
 14
 0
 13
 0
 5
 0
 4
 15
 3
 0
 0
 10

#include <bits/stdc++.h>
using namespace std;
namespace IO {
int read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
void write(int x, char sp) {
  char ch[20];
  int len = 0;
  if (x < 0) x = -x, putchar('-');
  do {
    ch[len++] = x % 10 + '0';
    x /= 10;
  } while (x);
  for (int i = len - 1; ~i; i--) putchar(ch[i]);
  putchar(sp);
}
void ckmin(int& x, int y) { x = x < y ? x : y; }
void ckmax(int& x, int y) { x = x > y ? x : y; }
}  // namespace IO
using namespace IO;
int n, U, ans, mn[30], val[30], sum[1 << 20], buc[800010], tim[800010][21],
    f[1 << 20];
const int bs = 400000;
char ch[30];
signed main() {
  n = read();
  U = (1 << n) - 1;
  for (int i = 1; i <= n; i++) {
    scanf("%s", ch + 1);
    memset(buc, 0, sizeof(buc));
    int m = 0, now = 0, len = strlen(ch + 1);
    for (int j = 1; j <= len; j++) {
      now += ch[j] == '(' ? 1 : -1;
      if (m > now) tim[bs + m][i] = buc[bs + m], --m;
      ++buc[bs + now];
    }
    tim[bs + m][i] = buc[bs + m];
    mn[i] = m;
    val[i] = now;
  }
  for (int i = 1; i <= U; i++)
    for (int j = 0; j < n; j++)
      if (i & (1 << j)) sum[i] += val[j + 1];
  memset(f, 0xc0, sizeof(f));
  f[0] = 0;
  for (int i = 0; i <= U; i++)
    for (int j = 1; j <= n; j++)
      if (!(i & (1 << j - 1)) && sum[i] + mn[j] >= 0)
        ckmax(f[i | (1 << j - 1)], f[i] + tim[bs - sum[i]][j]);
  for (int i = 0; i < +U; i++)
    if (sum[i] >= 0) {
      ckmax(ans, f[i]);
      for (int j = 1; j <= n; j++)
        if (!(i & (1 << j - 1))) ckmax(ans, f[i] + tim[bs - sum[i]][j]);
    }
  write(ans, '\n');
  return 0;
}
