#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool upmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T>
inline bool upmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
const long double eps = 1e-11;
const long double pi = acos(-1);
const int oo = 1 << 30;
const long long loo = 1ll << 62;
const int mods = 998244353;
const int MAXN = 600005;
const int INF = 0x3f3f3f3f;
inline int read() {
  int f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
int num = 0, cnt = 0;
map<long long, int> Map;
map<pair<int, int>, int> Ans;
set<pair<int, int> > Set[MAXN];
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
void add(int x, int y, int c) {
  int g = gcd(x, y);
  x /= g, y /= g;
  Ans[make_pair(x, y)] += c;
}
int query(int x, int y) {
  int g = gcd(x, y);
  x /= g, y /= g;
  return Ans[make_pair(x, y)];
}
int main() {
  int Case = read();
  while (Case--) {
    long long opt = read(), x = read(), y = read();
    if (opt == 1) {
      long long z = x * x + y * y;
      if (!Map.count(z)) Map[z] = ++cnt;
      z = Map[z];
      for (auto V : Set[z]) add(V.first + x, V.second + y, 2);
      Set[z].insert(make_pair(x, y)), add(x, y, 1), num++;
    } else if (opt == 2) {
      long long z = x * x + y * y;
      if (!Map.count(z)) Map[z] = ++cnt;
      z = Map[z];
      Set[z].erase(Set[z].find(make_pair(x, y))), add(x, y, -1), num--;
      for (auto V : Set[z]) add(V.first + x, V.second + y, -2);
    } else
      printf("%d\n", num - query(x, y));
  }
  return 0;
}

### Prompt

Your task is to create 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 <typename T>
inline bool upmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T>
inline bool upmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
const long double eps = 1e-11;
const long double pi = acos(-1);
const int oo = 1 << 30;
const long long loo = 1ll << 62;
const int mods = 998244353;
const int MAXN = 600005;
const int INF = 0x3f3f3f3f;
inline int read() {
  int f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
int num = 0, cnt = 0;
map<long long, int> Map;
map<pair<int, int>, int> Ans;
set<pair<int, int> > Set[MAXN];
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
void add(int x, int y, int c) {
  int g = gcd(x, y);
  x /= g, y /= g;
  Ans[make_pair(x, y)] += c;
}
int query(int x, int y) {
  int g = gcd(x, y);
  x /= g, y /= g;
  return Ans[make_pair(x, y)];
}
int main() {
  int Case = read();
  while (Case--) {
    long long opt = read(), x = read(), y = read();
    if (opt == 1) {
      long long z = x * x + y * y;
      if (!Map.count(z)) Map[z] = ++cnt;
      z = Map[z];
      for (auto V : Set[z]) add(V.first + x, V.second + y, 2);
      Set[z].insert(make_pair(x, y)), add(x, y, 1), num++;
    } else if (opt == 2) {
      long long z = x * x + y * y;
      if (!Map.count(z)) Map[z] = ++cnt;
      z = Map[z];
      Set[z].erase(Set[z].find(make_pair(x, y))), add(x, y, -1), num--;
      for (auto V : Set[z]) add(V.first + x, V.second + y, -2);
    } else
      printf("%d\n", num - query(x, y));
  }
  return 0;
}

```