#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int a[2005][2];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0), cout.precision(15);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i][0] >> a[i][1];
  }
  map<pair<pair<int, int>, pair<int, int> >, int> m;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int x1 = a[i][0], y1 = a[i][1];
      int x2 = a[j][0], y2 = a[j][1];
      if (y1 > y2) {
        swap(x2, x1);
        swap(y2, y1);
      }
      int mn = y2 - y1;
      int md = x2 - x1;
      int hn = y2 * x1 - y1 * x2;
      int hd = y2 - y1;
      if (md == 0) {
        m[make_pair(make_pair(1, 0), make_pair(x1, 1))]++;
        continue;
      }
      if (mn == 0) {
        m[make_pair(make_pair(0, 1), make_pair(y1, 1))]++;
        continue;
      }
      int mg = gcd(mn, abs(md));
      int hg = gcd(abs(hn), hd);
      mn /= mg;
      md /= mg;
      hn /= hg;
      hd /= hg;
      if (hn == 0) {
        m[make_pair(make_pair(mn, md), make_pair(0, 1))]++;
        continue;
      }
      m[make_pair(make_pair(mn, md), make_pair(hn, hd))]++;
    }
  }
  int answer = ((long long)(n * (n - 1)) * (n - 2)) / 6;
  for (map<pair<pair<int, int>, pair<int, int> >, int>::iterator it = m.begin();
       it != m.end(); it++) {
    int k = (*it).second;
    int nl = ceil(sqrt(2 * k));
    answer -= ((long long)(nl * (nl - 1)) * (nl - 2)) / 6;
  }
  cout << max(0, answer) << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int a[2005][2];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0), cout.precision(15);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i][0] >> a[i][1];
  }
  map<pair<pair<int, int>, pair<int, int> >, int> m;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      int x1 = a[i][0], y1 = a[i][1];
      int x2 = a[j][0], y2 = a[j][1];
      if (y1 > y2) {
        swap(x2, x1);
        swap(y2, y1);
      }
      int mn = y2 - y1;
      int md = x2 - x1;
      int hn = y2 * x1 - y1 * x2;
      int hd = y2 - y1;
      if (md == 0) {
        m[make_pair(make_pair(1, 0), make_pair(x1, 1))]++;
        continue;
      }
      if (mn == 0) {
        m[make_pair(make_pair(0, 1), make_pair(y1, 1))]++;
        continue;
      }
      int mg = gcd(mn, abs(md));
      int hg = gcd(abs(hn), hd);
      mn /= mg;
      md /= mg;
      hn /= hg;
      hd /= hg;
      if (hn == 0) {
        m[make_pair(make_pair(mn, md), make_pair(0, 1))]++;
        continue;
      }
      m[make_pair(make_pair(mn, md), make_pair(hn, hd))]++;
    }
  }
  int answer = ((long long)(n * (n - 1)) * (n - 2)) / 6;
  for (map<pair<pair<int, int>, pair<int, int> >, int>::iterator it = m.begin();
       it != m.end(); it++) {
    int k = (*it).second;
    int nl = ceil(sqrt(2 * k));
    answer -= ((long long)(nl * (nl - 1)) * (nl - 2)) / 6;
  }
  cout << max(0, answer) << endl;
  return 0;
}

```