#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
const int mxN = 2e3 + 1;
int n;
pii p[mxN];
int gcd(int a, int b) { return (b ? gcd(b, a % b) : a); }
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> p[i].first >> p[i].second;
  int cur = 0;
  for (int i = 0; i < n; ++i) {
    map<pair<int, int>, int> cnt;
    for (int j = i + 1; j < n; ++j) {
      int dy = p[i].second - p[j].second, dx = p[i].first - p[j].first;
      int neg = ((dy < 0) ^ (dx < 0)) ? -1 : 1;
      dy = abs(dy), dx = abs(dx);
      int GCD = gcd(dy, dx), num = dy * neg / GCD, den = dx / GCD;
      if (!num) den = abs(den);
      if (!den) num = abs(num);
      ++cnt[pii(num, den)];
    }
    for (auto p : cnt) cur += 1LL * p.second * (p.second - 1) / 2;
  }
  cout << 1LL * n * (n - 1) * (n - 2) / 6 - cur << '\n';
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
using pii = pair<int, int>;
const int mxN = 2e3 + 1;
int n;
pii p[mxN];
int gcd(int a, int b) { return (b ? gcd(b, a % b) : a); }
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> p[i].first >> p[i].second;
  int cur = 0;
  for (int i = 0; i < n; ++i) {
    map<pair<int, int>, int> cnt;
    for (int j = i + 1; j < n; ++j) {
      int dy = p[i].second - p[j].second, dx = p[i].first - p[j].first;
      int neg = ((dy < 0) ^ (dx < 0)) ? -1 : 1;
      dy = abs(dy), dx = abs(dx);
      int GCD = gcd(dy, dx), num = dy * neg / GCD, den = dx / GCD;
      if (!num) den = abs(den);
      if (!den) num = abs(num);
      ++cnt[pii(num, den)];
    }
    for (auto p : cnt) cur += 1LL * p.second * (p.second - 1) / 2;
  }
  cout << 1LL * n * (n - 1) * (n - 2) / 6 - cur << '\n';
}

```