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;
int n;
long long ans = 0;
struct Point {
  long long x, y;
  double thi;
  Point(){};
  Point(long long a, long long b) : x(a), y(b){};
  friend Point operator+(const Point &a, const Point &b) {
    return Point(a.x + b.x, a.y + b.y);
  }
  friend Point operator-(const Point &a, const Point &b) {
    return Point(a.x - b.x, a.y - b.y);
  }
  friend Point operator*(const Point &a, const double &r) {
    return Point(a.x * r, a.y * r);
  }
  friend Point operator~(const Point &a) { return Point(a.x * -1, a.y * -1); }
  bool operator<(const Point &X) const { return thi < X.thi; }
} p[307];
ostream &operator<<(ostream &o, Point x) { return o << x.x << " " << x.y; }
long long cross(const Point &a, const Point &b) {
  return a.x * b.y - a.y * b.x;
}
int f[2007][2007][7];
struct node {
  double k;
  int s, t;
  node(){};
  node(double k, int s, int t) : k(k), s(s), t(t){};
  bool operator<(const node &X) const { return k < X.k; }
};
vector<node> v;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld", &p[i].x, &p[i].y);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      v.push_back(node(atan2(p[j].y - p[i].y, p[j].x - p[i].x), i, j));
      v.push_back(node(atan2(p[i].y - p[j].y, p[i].x - p[j].x), j, i));
    }
  }
  sort(v.begin(), v.end());
  for (auto y : v) {
    f[y.s][y.t][0]++;
    for (int i = 1; i <= n; i++) {
      f[i][y.t][1] += f[i][y.s][0];
      f[i][y.t][2] += f[i][y.s][1];
      f[i][y.t][3] += f[i][y.s][2];
      f[i][y.t][4] += f[i][y.s][3];
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += f[i][i][4];
  }
  printf("%lld\n", ans);
  return 0;
}
