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;
typedef class Point {
 public:
  int x, y;
  Point() {}
  Point(int x, int y) : x(x), y(y) {}
} Point;
typedef class Line {
 public:
  Point v;
  int a, b;
  double ang;
  Line(Point v, int a, int b) : v(v), a(a), b(b) { ang = atan2(v.y, v.x); }
  bool operator<(Line b) const { return ang < b.ang; }
} Line;
Point operator-(Point a, Point b) { return Point(a.x - b.x, a.y - b.y); }
const int N = 303;
int n;
Point P[N];
long long dp[N][5][N];
vector<Line> E;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &P[i].x, &P[i].y);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i ^ j) {
        E.emplace_back(P[j] - P[i], i, j);
      }
    }
  }
  sort(E.begin(), E.end());
  for (auto e : E) {
    int a = e.a, b = e.b;
    dp[b][0][a] = 1;
    for (int t = 0; t < 4; t++) {
      for (int c = 1; c <= n; c++) {
        dp[b][t + 1][c] += dp[a][t][c];
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += dp[i][4][i];
  printf("%lld\n", ans);
  return 0;
}
