#include <bits/stdc++.h>
using namespace std;
const int M = 2e3 + 10;
int n;
struct Point {
  int x, y;
  Point() {}
  Point(const int &x, const int &y) : x(x), y(y) {}
  int dim() const { return x < 0 || x == 0 && y < 0; }
  void in() { cin >> x >> y; }
} a[M];
Point b[M];
long long tot, cut;
int operator*(const Point &a, const Point &b) { return a.x * b.y - a.y * b.x; }
bool operator<(const Point &a, const Point &b) {
  if (a.dim() == b.dim()) {
    return a * b > 0;
  } else {
    return a.dim() > b.dim();
  }
}
bool operator==(const Point &a, const Point &b) { return !(a < b) && !(b < a); }
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) a[i].in();
  tot = 1ll * n * (n - 1) * (n - 2) / 6;
  for (int i = 0; i < n; i++) {
    for (int j = 0, N = 0; j < n; j++) {
      if (i == j) continue;
      b[N] = Point(a[j].x - a[i].x, a[j].y - a[i].y);
      N++;
    }
    sort(b, b + n - 1);
    int tmp = 0;
    for (int j = 1; j < n - 1; j++) {
      if (b[j] == b[j - 1])
        tmp++;
      else {
        cut += 1ll * (tmp + 1) * tmp / 2;
        tmp = 0;
      }
    }
    cut += 1ll * (tmp + 1) * tmp / 2;
  }
  cout << tot - cut / 2 << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

Vanya got bored and he painted n distinct points on the plane. After that he connected all the points pairwise and saw that as a result many triangles were formed with vertices in the painted points. He asks you to count the number of the formed triangles with the non-zero area.

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of the points painted on the plane. 

Next n lines contain two integers each xi, yi ( - 100 ≤ xi, yi ≤ 100) — the coordinates of the i-th point. It is guaranteed that no two given points coincide.

Output

In the first line print an integer — the number of triangles with the non-zero area among the painted points.

Examples

Input

4
0 0
1 1
2 0
2 2


Output

3


Input

3
0 0
1 1
2 0


Output

1


Input

1
1 1


Output

0

Note

Note to the first sample test. There are 3 triangles formed: (0, 0) - (1, 1) - (2, 0); (0, 0) - (2, 2) - (2, 0); (1, 1) - (2, 2) - (2, 0).

Note to the second sample test. There is 1 triangle formed: (0, 0) - (1, 1) - (2, 0).

Note to the third sample test. A single point doesn't form a single triangle.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int M = 2e3 + 10;
int n;
struct Point {
  int x, y;
  Point() {}
  Point(const int &x, const int &y) : x(x), y(y) {}
  int dim() const { return x < 0 || x == 0 && y < 0; }
  void in() { cin >> x >> y; }
} a[M];
Point b[M];
long long tot, cut;
int operator*(const Point &a, const Point &b) { return a.x * b.y - a.y * b.x; }
bool operator<(const Point &a, const Point &b) {
  if (a.dim() == b.dim()) {
    return a * b > 0;
  } else {
    return a.dim() > b.dim();
  }
}
bool operator==(const Point &a, const Point &b) { return !(a < b) && !(b < a); }
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) a[i].in();
  tot = 1ll * n * (n - 1) * (n - 2) / 6;
  for (int i = 0; i < n; i++) {
    for (int j = 0, N = 0; j < n; j++) {
      if (i == j) continue;
      b[N] = Point(a[j].x - a[i].x, a[j].y - a[i].y);
      N++;
    }
    sort(b, b + n - 1);
    int tmp = 0;
    for (int j = 1; j < n - 1; j++) {
      if (b[j] == b[j - 1])
        tmp++;
      else {
        cut += 1ll * (tmp + 1) * tmp / 2;
        tmp = 0;
      }
    }
    cut += 1ll * (tmp + 1) * tmp / 2;
  }
  cout << tot - cut / 2 << endl;
  return 0;
}

```