#include <bits/stdc++.h>
using namespace std;
template <class U, class T>
void Max(U &first, T second) {
  if (first < second) first = second;
}
template <class U, class T>
void Min(U &first, T second) {
  if (first > second) first = second;
}
template <class T>
void add(int &a, T b) {
  a = (a + b) % 1000000007;
}
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % 1000000007;
    a = 1LL * a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int pow(int a, int b, int c) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % c;
    a = 1LL * a * a % c;
    b >>= 1;
  }
  return ans;
}
map<pair<int, int>, int> w;
map<long long, vector<pair<int, int>>> q;
void add(long long l, pair<int, int> e) {
  for (int i = 0; i < q[l].size(); i++) {
    pair<int, int> &o = q[l][i];
    int first = e.first + o.first, second = e.second + o.second;
    int d = gcd(first, second);
    first /= d, second /= d;
    w[make_pair(first, second)] += 2;
  }
  q[l].push_back(e);
}
void del(long long l, pair<int, int> e) {
  for (int i = 0; i < q[l].size(); i++)
    if (q[l][i] == e) {
      swap(q[l][i], q[l].back());
      q[l].pop_back();
      break;
    }
  for (int i = 0; i < q[l].size(); i++) {
    pair<int, int> &o = q[l][i];
    int first = e.first + o.first, second = e.second + o.second;
    int d = gcd(first, second);
    first /= d, second /= d;
    w[make_pair(first, second)] -= 2;
  }
}
int main() {
  int ca = 0, T, k, i, j, m = 0, K, n = 0;
  scanf("%d", &T);
  int first, second;
  n = 0;
  while (T--) {
    scanf("%d%d%d", &k, &first, &second);
    int d = gcd(first, second);
    pair<int, int> e = make_pair(first / d, second / d);
    if (k == 1) {
      long long l = 1LL * first * first + 1LL * second * second;
      add(l, {first, second});
      w[e]++;
      n++;
    } else if (k == 2) {
      long long l = 1LL * first * first + 1LL * second * second;
      del(l, {first, second});
      w[e]--;
      n--;
    } else {
      if (w.count(e))
        K = w[e];
      else
        K = 0;
      printf("%d\n", n - K);
    }
  }
}

### Prompt

Generate a cpp solution to the following problem:

Consider a set of points A, initially it is empty. There are three types of queries: 

  1. Insert a point (x_i, y_i) to A. It is guaranteed that this point does not belong to A at this moment. 
  2. Remove a point (x_i, y_i) from A. It is guaranteed that this point belongs to A at this moment. 
  3. Given a point (x_i, y_i), calculate the minimum number of points required to add to A to make A symmetrical with respect to the line containing points (0, 0) and (x_i, y_i). Note that these points are not actually added to A, i.e. these queries are independent from each other. 

Input

The first line contains a single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.

Each of the following q lines describes a query and contains three integers t_i, x_i and y_i ( t_i ∈ \{1, 2, 3\}, 1 ≤ x_i, y_i ≤ 112 904) — the type of the query and the coordinates of the point. Type 1 is addition of the point, type 2 is removal of the point, type 3 is the query to compute the minimum number of points required to make A symmetrical. 

It is guaranteed that there are no more than 10^5 queries of type 3 and no more than 10^5 queries having type 1 or 2.

Output

For each query of the third type output a line with a single integer — the answer to this query.

Examples

Input

12
1 1 6
1 6 1
1 5 5
1 2 3
3 4 4
1 3 2
3 7 7
2 2 3
2 6 1
3 8 8
2 5 5
3 1 1


Output

1
0
2
2


Input

6
1 1 2
3 1 1
1 1 1
3 2 2
2 1 1
3 2 4


Output

1
1
0

Note

The first example is shown on the picture below.

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class U, class T>
void Max(U &first, T second) {
  if (first < second) first = second;
}
template <class U, class T>
void Min(U &first, T second) {
  if (first > second) first = second;
}
template <class T>
void add(int &a, T b) {
  a = (a + b) % 1000000007;
}
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % 1000000007;
    a = 1LL * a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int pow(int a, int b, int c) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1LL * ans * a % c;
    a = 1LL * a * a % c;
    b >>= 1;
  }
  return ans;
}
map<pair<int, int>, int> w;
map<long long, vector<pair<int, int>>> q;
void add(long long l, pair<int, int> e) {
  for (int i = 0; i < q[l].size(); i++) {
    pair<int, int> &o = q[l][i];
    int first = e.first + o.first, second = e.second + o.second;
    int d = gcd(first, second);
    first /= d, second /= d;
    w[make_pair(first, second)] += 2;
  }
  q[l].push_back(e);
}
void del(long long l, pair<int, int> e) {
  for (int i = 0; i < q[l].size(); i++)
    if (q[l][i] == e) {
      swap(q[l][i], q[l].back());
      q[l].pop_back();
      break;
    }
  for (int i = 0; i < q[l].size(); i++) {
    pair<int, int> &o = q[l][i];
    int first = e.first + o.first, second = e.second + o.second;
    int d = gcd(first, second);
    first /= d, second /= d;
    w[make_pair(first, second)] -= 2;
  }
}
int main() {
  int ca = 0, T, k, i, j, m = 0, K, n = 0;
  scanf("%d", &T);
  int first, second;
  n = 0;
  while (T--) {
    scanf("%d%d%d", &k, &first, &second);
    int d = gcd(first, second);
    pair<int, int> e = make_pair(first / d, second / d);
    if (k == 1) {
      long long l = 1LL * first * first + 1LL * second * second;
      add(l, {first, second});
      w[e]++;
      n++;
    } else if (k == 2) {
      long long l = 1LL * first * first + 1LL * second * second;
      del(l, {first, second});
      w[e]--;
      n--;
    } else {
      if (w.count(e))
        K = w[e];
      else
        K = 0;
      printf("%d\n", n - K);
    }
  }
}

```