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;
const int N = 1005;
const int Q = 1e9 + 7;
struct Point {
  int x, y;
  Point(int _x = 0, int _y = 0) : x(_x), y(_y) {}
  const int sign() const { return y < 0 || y == 0 && x < 0; }
  Point operator-(const Point &R) { return Point(x - R.x, y - R.y); }
  long long operator^(const Point &R) const {
    return (long long)x * R.y - (long long)y * R.x;
  }
  bool operator<(const Point &R) const {
    if (sign() != R.sign()) {
      return sign() < R.sign();
    }
    return (*this ^ R) > 0;
  }
};
inline void add(long long &A, long long B) { A += B; }
Point p[N];
int n;
long long f[6][N][N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d%d", &p[i].x, &p[i].y);
    f[0][i][i] = 1;
  }
  vector<tuple<Point, int, int>> v;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < i; ++j) {
      v.emplace_back(p[j] - p[i], i, j);
      v.emplace_back(p[i] - p[j], j, i);
    }
  }
  sort(v.begin(), v.end());
  for (auto &it : v) {
    int x = get<1>(it);
    int y = get<2>(it);
    for (int i = 5; i > 0; --i) {
      for (int j = 0; j < n; ++j) {
        add(f[i][j][y], f[i - 1][j][x]);
      }
    }
  }
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    add(res, f[5][i][i]);
  }
  cout << res << endl;
}
