#include <bits/stdc++.h>
using namespace std;
const int N = 2020;
const int C = 204;
int gcd(int x, int y) { return (y == 0 ? x : gcd(y, x % y)); }
int x[N], y[N];
int a[C][2 * C];
int n;
long long ans;
void solve(int id) {
  memset(a, 0, sizeof a);
  for (int i = id + 1; i < n; i++) {
    int xx = x[i] - x[id];
    int yy = y[i] - y[id];
    if (xx < 0) {
      xx *= -1;
      yy *= -1;
    }
    if (xx == 0 && yy < 0) yy *= -1;
    int g = gcd(xx, abs(yy));
    xx /= g;
    yy /= g;
    ans -= a[xx][yy];
    a[xx][yy]++;
  }
  return;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
  ans = (long long)n * (n - 1) * (n - 2) / 6LL;
  for (int i = 0; i < n; i++) solve(i);
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution 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 N = 2020;
const int C = 204;
int gcd(int x, int y) { return (y == 0 ? x : gcd(y, x % y)); }
int x[N], y[N];
int a[C][2 * C];
int n;
long long ans;
void solve(int id) {
  memset(a, 0, sizeof a);
  for (int i = id + 1; i < n; i++) {
    int xx = x[i] - x[id];
    int yy = y[i] - y[id];
    if (xx < 0) {
      xx *= -1;
      yy *= -1;
    }
    if (xx == 0 && yy < 0) yy *= -1;
    int g = gcd(xx, abs(yy));
    xx /= g;
    yy /= g;
    ans -= a[xx][yy];
    a[xx][yy]++;
  }
  return;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
  ans = (long long)n * (n - 1) * (n - 2) / 6LL;
  for (int i = 0; i < n; i++) solve(i);
  cout << ans << endl;
  return 0;
}

```