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;
int n;
const int inf = 1e9, N = 25, M = 4e5 + 5, S = 4e6 + 5;
struct point {
  int w, sum;
  vector<int> t;
} a[N];
inline int read() {
  int F = 1, ANS = 0;
  char C = getchar();
  while (C < '0' || C > '9') {
    if (C == '-') F = -1;
    C = getchar();
  }
  while (C >= '0' && C <= '9') {
    ANS = ANS * 10 + C - '0';
    C = getchar();
  }
  return F * ANS;
}
char s[M];
int dp[S], ans;
inline char readchar() {
  char C = getchar();
  while (C < 33 || C > 126) C = getchar();
  return C;
}
inline int raed() { return read(); }
int main() {
  n = read();
  for (int i = 0; i < n; i++) {
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    int tot = 0, res = 0;
    for (int j = 0; j <= len; j++) a[i].t.push_back(0);
    for (int j = 1; j <= len; j++) {
      if (s[j] == '(')
        tot++;
      else
        tot--;
      if (tot <= res) {
        res = tot;
        a[i].t[-tot]++;
      }
    }
    a[i].w = -res;
    a[i].sum = tot;
  }
  for (int i = 1; i <= (1 << n) - 1; i++) dp[i] = -inf;
  dp[S] = 0;
  for (int i = 1; i <= (1 << n) - 1; i++) {
    int sum = 0;
    for (int j = 0; j < n; j++) {
      if (((1 << j) & i) == 0) continue;
      sum += a[j].sum;
    }
    for (int j = 0; j < n; j++) {
      if (((1 << j) & i) == 0) continue;
      sum -= a[j].sum;
      int k = ((1 << j) ^ i);
      if (sum < 0) {
        sum += a[j].sum;
        continue;
      }
      if (dp[k] < 0) {
        sum += a[j].sum;
        continue;
      }
      if (sum > a[j].w) {
        dp[i] = max(dp[i], dp[k]);
        sum += a[j].sum;
        continue;
      }
      if (sum == a[j].w) {
        dp[i] = max(dp[i], dp[k] + a[j].t[sum]);
        sum += a[j].sum;
        continue;
      }
      ans = max(ans, dp[k] + a[j].t[sum]);
      sum += a[j].sum;
    }
    ans = max(ans, dp[i]);
  }
  cout << ans << endl;
  return 0;
}
