You are given a set of n points in a 2D plane. No three points are collinear.

A pentagram is a set of 5 points A,B,C,D,E that can be arranged as follows. <image> Note the length of the line segments don't matter, only that those particular intersections exist.

Count the number of ways to choose 5 points from the given set that form a pentagram.

Input

The first line contains an integer n (5 ≤ n ≤ 300) — the number of points.

Each of the next n lines contains two integers x_i, y_i (-10^6 ≤ x_i,y_i ≤ 10^6) — the coordinates of the i-th point. It is guaranteed that no three points are collinear.

Output

Print a single integer, the number of sets of 5 points that form a pentagram.

Examples

Input


5
0 0
0 2
2 0
2 2
1 3


Output


1


Input


5
0 0
4 0
0 4
4 4
2 3


Output


0


Input


10
841746 527518
595261 331297
-946901 129987
670374 -140388
-684770 309555
-302589 415564
-387435 613331
-624940 -95922
945847 -199224
24636 -565799


Output


85

Note

A picture of the first sample: <image> A picture of the second sample: <image> A picture of the third sample: <image>
#include <bits/stdc++.h>
using namespace std;
struct hh {
  int x, y;
  bool operator<(hh a) const { return x < a.x; }
} sk[305];
int n, id[305][305], sum[305][305], ac[305][305], ls[305], rt;
double w[305][305];
bool cmp(int x, int y) { return w[rt][x] < w[rt][y]; }
int query1(int i, int a, int b) {
  int l = 1, r = n - 1, pr1 = n - 1, pr2 = id[i][a], ans = 0;
  double s = atan2(sk[a].y - sk[b].y, sk[a].x - sk[b].x);
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (w[a][ac[a][mid]] < s)
      l = mid + 1, pr1 = mid;
    else
      r = mid - 1;
  }
  if (pr1 >= pr2)
    ans = sum[a][pr1] - sum[a][pr2];
  else
    ans = sum[a][n - 1] - sum[a][pr2] + sum[a][pr1];
  return max(0, ans);
}
int query2(int i, int a, int b) {
  int l = 1, r = n - 1, pr1 = 1, pr2 = id[i][a], ans = 0;
  double s = atan2(sk[a].y - sk[b].y, sk[a].x - sk[b].x);
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (w[a][ac[a][mid]] > s)
      r = mid - 1, pr1 = mid;
    else
      l = mid + 1;
  }
  if (pr2 >= pr1)
    ans = sum[a][pr2 - 1] - sum[a][pr1 - 1];
  else
    ans = sum[a][n - 1] - sum[a][pr1 - 1] + sum[a][pr2 - 1];
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &sk[i].x, &sk[i].y);
  sort(sk + 1, sk + n + 1);
  for (int i = 1; i <= n; ++i) {
    int tot = 0;
    rt = i;
    for (int j = 1; j <= n; ++j)
      if (j != i)
        ls[++tot] = j, w[i][j] = atan2(sk[j].y - sk[i].y, sk[j].x - sk[i].x);
    sort(ls + 1, ls + tot + 1, cmp);
    for (int j = 1; j <= tot; ++j)
      ac[i][j] = ls[j], id[ls[j]][i] = j, sum[i][j] = j;
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n - 1; ++j) {
      int a = ac[i][j];
      if (a <= i) continue;
      for (int k = j + 1; k <= n - 1; ++k) {
        int b = ac[i][k];
        if (b <= i) continue;
        int w1 = query1(i, a, b), w2 = query2(i, b, a);
        ans += w1 * w2;
      }
    }
    for (int j = i + 1; j <= n; ++j)
      for (int k = id[i][j]; k <= n - 1; ++k) sum[j][k]--;
  }
  printf("%lld", ans);
}
