#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long mod2 = 998244353;
long long mod3 = 1000003;
long long mod4 = 998244853;
long long inf = 1000000000000000000;
double pi = acos(-1.0);
int dh[4] = {1, -1, 0, 0};
int dw[4] = {0, 0, 1, -1};
int ddh[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddw[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
long long lmax(long long a, long long b) {
  if (a < b)
    return b;
  else
    return a;
}
long long lmin(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long Pow(long long n, long long k) {
  long long ret = 1;
  long long now = n;
  while (k > 0) {
    if (k & 1) ret *= now;
    now *= now;
    k /= 2;
  }
  return ret;
}
long long beki(long long n, long long k, long long md) {
  long long ret = 1;
  long long now = n;
  while (k > 0) {
    if (k % 2 == 1) {
      ret *= now;
      ret %= md;
    }
    now *= now;
    now %= md;
    k /= 2;
  }
  return ret;
}
long long gyaku(long long n, long long md) { return beki(n, md - 2, md); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  int a[n];
  int sum[500001];
  int sum2[500001];
  fill(sum, sum + 500001, 0);
  fill(sum2, sum2 + 500001, 0);
  for (long long i = 0; i < n; i++) {
    string s;
    cin >> s;
    a[i] = 0;
    int m = s.size();
    if (s[0] == ')') {
      if (s[m - 1] == '(') {
        a[i] = 1000000000;
        continue;
      }
      int cnt = 0;
      for (long long j = m - 1; j >= 0; j--) {
        if (s[j] == '(')
          cnt--;
        else
          cnt++;
        if (cnt < 0) {
          a[i] = 1000000000;
          break;
        }
      }
      if (a[i] != 0) continue;
      a[i] = -cnt;
      sum2[-a[i]]++;
    } else if (s[m - 1] == '(') {
      int cnt = 0;
      for (long long j = 0; j < m; j++) {
        if (s[j] == ')')
          cnt--;
        else
          cnt++;
        if (cnt < 0) {
          a[i] = 1000000000;
          break;
        }
      }
      if (a[i] != 0) continue;
      a[i] = cnt;
      sum[a[i]]++;
    } else {
      int cnt = 0;
      int mi = 0, ma = 0;
      for (long long j = 0; j < m; j++) {
        if (s[j] == ')')
          cnt--;
        else
          cnt++;
        mi = min(mi, cnt);
      }
      cnt = 0;
      for (long long j = m - 1; j >= 0; j--) {
        if (s[j] == '(')
          cnt--;
        else
          cnt++;
        ma = min(ma, cnt);
      }
      if (ma != 0 && mi != 0) continue;
      if (ma < 0)
        sum[-ma]++;
      else
        sum2[-mi]++;
    }
  }
  int ans = 0;
  for (long long i = 1; i < 500001; i++) ans += min(sum[i], sum2[i]);
  ans += (sum[0] + sum2[0]) / 2;
  cout << ans << endl;
}

### Prompt

Please create a solution in Cpp to the following problem:

One day, Yuhao came across a problem about checking if some bracket sequences are correct bracket sequences.

A bracket sequence is any non-empty sequence of opening and closing parentheses. A bracket sequence is called a correct bracket sequence if it's possible to obtain a correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, the sequences "(())()", "()" and "(()(()))" are correct, while the bracket sequences ")(", "(()" and "(()))(" are not correct.

Yuhao found this problem too simple for him so he decided to make the problem harder. You are given many (not necessarily correct) bracket sequences. The task is to connect some of them into ordered pairs so that each bracket sequence occurs in at most one pair and the concatenation of the bracket sequences in each pair is a correct bracket sequence. The goal is to create as many pairs as possible.

This problem unfortunately turned out to be too difficult for Yuhao. Can you help him and solve it?

Input

The first line contains one integer n (1 ≤ n ≤ 10^5) — the number of bracket sequences.

Each of the following n lines contains one bracket sequence — a non-empty string which consists only of characters "(" and ")".

The sum of lengths of all bracket sequences in the input is at most 5 ⋅ 10^5.

Note that a bracket sequence may appear in the input multiple times. In this case, you can use each copy of the sequence separately. Also note that the order in which strings appear in the input doesn't matter.

Output

Print a single integer — the maximum number of pairs which can be made, adhering to the conditions in the statement.

Examples

Input


7
)())
)
((
((
(
)
)


Output


2


Input


4
(
((
(((
(())


Output


0


Input


2
(())
()


Output


1

Note

In the first example, it's optimal to construct two pairs: "(( )())" and "( )".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long mod2 = 998244353;
long long mod3 = 1000003;
long long mod4 = 998244853;
long long inf = 1000000000000000000;
double pi = acos(-1.0);
int dh[4] = {1, -1, 0, 0};
int dw[4] = {0, 0, 1, -1};
int ddh[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddw[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
long long lmax(long long a, long long b) {
  if (a < b)
    return b;
  else
    return a;
}
long long lmin(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long Pow(long long n, long long k) {
  long long ret = 1;
  long long now = n;
  while (k > 0) {
    if (k & 1) ret *= now;
    now *= now;
    k /= 2;
  }
  return ret;
}
long long beki(long long n, long long k, long long md) {
  long long ret = 1;
  long long now = n;
  while (k > 0) {
    if (k % 2 == 1) {
      ret *= now;
      ret %= md;
    }
    now *= now;
    now %= md;
    k /= 2;
  }
  return ret;
}
long long gyaku(long long n, long long md) { return beki(n, md - 2, md); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  int a[n];
  int sum[500001];
  int sum2[500001];
  fill(sum, sum + 500001, 0);
  fill(sum2, sum2 + 500001, 0);
  for (long long i = 0; i < n; i++) {
    string s;
    cin >> s;
    a[i] = 0;
    int m = s.size();
    if (s[0] == ')') {
      if (s[m - 1] == '(') {
        a[i] = 1000000000;
        continue;
      }
      int cnt = 0;
      for (long long j = m - 1; j >= 0; j--) {
        if (s[j] == '(')
          cnt--;
        else
          cnt++;
        if (cnt < 0) {
          a[i] = 1000000000;
          break;
        }
      }
      if (a[i] != 0) continue;
      a[i] = -cnt;
      sum2[-a[i]]++;
    } else if (s[m - 1] == '(') {
      int cnt = 0;
      for (long long j = 0; j < m; j++) {
        if (s[j] == ')')
          cnt--;
        else
          cnt++;
        if (cnt < 0) {
          a[i] = 1000000000;
          break;
        }
      }
      if (a[i] != 0) continue;
      a[i] = cnt;
      sum[a[i]]++;
    } else {
      int cnt = 0;
      int mi = 0, ma = 0;
      for (long long j = 0; j < m; j++) {
        if (s[j] == ')')
          cnt--;
        else
          cnt++;
        mi = min(mi, cnt);
      }
      cnt = 0;
      for (long long j = m - 1; j >= 0; j--) {
        if (s[j] == '(')
          cnt--;
        else
          cnt++;
        ma = min(ma, cnt);
      }
      if (ma != 0 && mi != 0) continue;
      if (ma < 0)
        sum[-ma]++;
      else
        sum2[-mi]++;
    }
  }
  int ans = 0;
  for (long long i = 1; i < 500001; i++) ans += min(sum[i], sum2[i]);
  ans += (sum[0] + sum2[0]) / 2;
  cout << ans << endl;
}

```