#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
  if (b) return gcd(b, a % b);
  return a;
}
int main() {
  int n;
  scanf("%d", &n);
  int p[n][2];
  int i, j;
  for (i = 0; i < n; i++) {
    scanf("%d %d", &p[i][0], &p[i][1]);
  }
  int c1, c2, d;
  long long res = 0, sum;
  for (i = 0; i < n; i++) {
    map<pair<int, int>, int> mm;
    for (j = 0; j < i; j++) {
      c1 = p[i][0] - p[j][0];
      c2 = p[i][1] - p[j][1];
      if (c1 == 0)
        mm[pair<int, int>(0, 1)]++;
      else if (c2 == 0)
        mm[pair<int, int>(1, 0)]++;
      else {
        d = gcd(max(abs(c1), abs(c2)), min(abs(c1), abs(c2)));
        c1 /= d;
        c2 /= d;
        if (c1 < 0) {
          c1 = -c1;
          c2 = -c2;
        }
        mm[pair<int, int>(c1, c2)]++;
      }
    }
    for (j = i + 1; j < n; j++) {
      c1 = p[i][0] - p[j][0];
      c2 = p[i][1] - p[j][1];
      if (c1 == 0)
        mm[pair<int, int>(0, 1)]++;
      else if (c2 == 0)
        mm[pair<int, int>(1, 0)]++;
      else {
        d = gcd(max(abs(c1), abs(c2)), min(abs(c1), abs(c2)));
        c1 /= d;
        c2 /= d;
        if (c1 < 0) {
          c1 = -c1;
          c2 = -c2;
        }
        mm[pair<int, int>(c1, c2)]++;
      }
    }
    sum = 0;
    map<pair<int, int>, int>::iterator it;
    for (it = mm.begin(); it != mm.end(); it++) {
      sum += it->second * (n - it->second - 1);
    }
    res += sum / 2;
  }
  res /= 3;
  printf("%lld\n", res);
  return 0;
}

### Prompt

Please formulate 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) {
  if (b) return gcd(b, a % b);
  return a;
}
int main() {
  int n;
  scanf("%d", &n);
  int p[n][2];
  int i, j;
  for (i = 0; i < n; i++) {
    scanf("%d %d", &p[i][0], &p[i][1]);
  }
  int c1, c2, d;
  long long res = 0, sum;
  for (i = 0; i < n; i++) {
    map<pair<int, int>, int> mm;
    for (j = 0; j < i; j++) {
      c1 = p[i][0] - p[j][0];
      c2 = p[i][1] - p[j][1];
      if (c1 == 0)
        mm[pair<int, int>(0, 1)]++;
      else if (c2 == 0)
        mm[pair<int, int>(1, 0)]++;
      else {
        d = gcd(max(abs(c1), abs(c2)), min(abs(c1), abs(c2)));
        c1 /= d;
        c2 /= d;
        if (c1 < 0) {
          c1 = -c1;
          c2 = -c2;
        }
        mm[pair<int, int>(c1, c2)]++;
      }
    }
    for (j = i + 1; j < n; j++) {
      c1 = p[i][0] - p[j][0];
      c2 = p[i][1] - p[j][1];
      if (c1 == 0)
        mm[pair<int, int>(0, 1)]++;
      else if (c2 == 0)
        mm[pair<int, int>(1, 0)]++;
      else {
        d = gcd(max(abs(c1), abs(c2)), min(abs(c1), abs(c2)));
        c1 /= d;
        c2 /= d;
        if (c1 < 0) {
          c1 = -c1;
          c2 = -c2;
        }
        mm[pair<int, int>(c1, c2)]++;
      }
    }
    sum = 0;
    map<pair<int, int>, int>::iterator it;
    for (it = mm.begin(); it != mm.end(); it++) {
      sum += it->second * (n - it->second - 1);
    }
    res += sum / 2;
  }
  res /= 3;
  printf("%lld\n", res);
  return 0;
}

```