#include <bits/stdc++.h>
using namespace std;
struct zz {
  double x;
  double y;
} p[2010];
bool cmp1(zz a, zz b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
struct ss {
  double k;
  double b;
} pp[2010 * 2010];
bool cmp2(ss a, ss b) {
  if (a.k == b.k) return a.b < b.b;
  return a.k < b.k;
}
long long C(int x, int k) {
  int i, j;
  long long s = 1;
  for (i = x, j = 1; i > x - k; i--, j++) s *= i, s /= j;
  return s;
}
int main() {
  int n, i, j, k;
  while (scanf("%d", &n) != EOF) {
    for (i = 0; i < n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
    sort(p, p + n, cmp1);
    k = 0;
    double x, y;
    for (i = 0; i < n - 1; i++) {
      for (j = i + 1; j < n; j++) {
        x = (p[j].x - p[i].x);
        y = (p[j].y - p[i].y);
        if (x == 0) {
          pp[k].k = 1000.0;
          pp[k].b = p[j].x;
        } else {
          pp[k].k = y / x;
          pp[k].b = p[i].y - pp[k].k * p[i].x;
        }
        k++;
      }
    }
    sort(pp, pp + k, cmp2);
    int cnt = 1, kk;
    long long num = C(n, 3);
    for (i = 0; i < k; i++) {
      if (fabs(pp[i].k - pp[i + 1].k) <= 1e-7 &&
          fabs(pp[i].b - pp[i + 1].b) <= 1e-7)
        cnt++;
      else {
        kk = (sqrt(1 + 8 * cnt) + 1) / 2;
        num -= C(kk, 3);
        cnt = 1;
      }
    }
    printf("%lld\n", num);
  }
  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;
struct zz {
  double x;
  double y;
} p[2010];
bool cmp1(zz a, zz b) {
  if (a.x == b.x) return a.y < b.y;
  return a.x < b.x;
}
struct ss {
  double k;
  double b;
} pp[2010 * 2010];
bool cmp2(ss a, ss b) {
  if (a.k == b.k) return a.b < b.b;
  return a.k < b.k;
}
long long C(int x, int k) {
  int i, j;
  long long s = 1;
  for (i = x, j = 1; i > x - k; i--, j++) s *= i, s /= j;
  return s;
}
int main() {
  int n, i, j, k;
  while (scanf("%d", &n) != EOF) {
    for (i = 0; i < n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
    sort(p, p + n, cmp1);
    k = 0;
    double x, y;
    for (i = 0; i < n - 1; i++) {
      for (j = i + 1; j < n; j++) {
        x = (p[j].x - p[i].x);
        y = (p[j].y - p[i].y);
        if (x == 0) {
          pp[k].k = 1000.0;
          pp[k].b = p[j].x;
        } else {
          pp[k].k = y / x;
          pp[k].b = p[i].y - pp[k].k * p[i].x;
        }
        k++;
      }
    }
    sort(pp, pp + k, cmp2);
    int cnt = 1, kk;
    long long num = C(n, 3);
    for (i = 0; i < k; i++) {
      if (fabs(pp[i].k - pp[i + 1].k) <= 1e-7 &&
          fabs(pp[i].b - pp[i + 1].b) <= 1e-7)
        cnt++;
      else {
        kk = (sqrt(1 + 8 * cnt) + 1) / 2;
        num -= C(kk, 3);
        cnt = 1;
      }
    }
    printf("%lld\n", num);
  }
  return 0;
}

```